Quadratic Inequality Calculator (Solution Set, Discriminant Cases and Graph)

Enter the coefficients a, b and c of the left side, ax² + bx + c, and the sign that compares it with 0. The inequality below is linked to the input fields, so you can also edit the coefficients directly in it.

Coefficient of x², a

Coefficient of x, b

Constant term c

Inequality sign (right side is 0)

Decimals, negative numbers and fractions such as 3/4 can be used. A blank a is treated as 1 (x² means 1x²), and a blank b or c is treated as 0. If a is 0, it is solved as a linear inequality.

Result and graph

Enter the coefficients a, b and c and the inequality sign in the fields on the left and press "Calculate". The solution set, the steps and a graph will appear here.

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What you can do on this page
What is this calculation used for?
How to Use
Formulas and graphs
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Find the solution set of a quadratic inequality such as $ax^{2}+bx+c>0$ just by entering the coefficients and choosing the sign. All four signs, $>$, $\ge$, $<$ and $\le$, are supported
The steps split into cases by the sign of the discriminant $D=b^{2}-4ac$, line by line, in the same order as a textbook. Besides "outside the two roots" and "between the two roots", it also handles tricky answers such as $x \neq \alpha$, no solution and all real numbers
When the coefficient of $x^{2}$ is negative, the steps also show multiplying both sides by $-1$ and flipping the inequality sign
Boundary values are shown as exact values with simplified radicals, such as $\dfrac{1+\sqrt{5}}{2}$, with decimals added when they are irrational. Coefficients can be decimals, negative numbers or fractions such as 3/4
A color-coded graph of the parabola and the solution set lets you see that "the solution of $y>0$ is the range of $x$ where the graph is above the $x$-axis"
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

The right side is fixed at 0. If there is an expression on the right, as in $x^{2}+2x > 3x+1$, move everything to the left first (here, enter $x^{2}-x-1>0$). If you enter 0 for the coefficient $a$ of $x^{2}$, it is not a quadratic inequality, so it is solved as the linear inequality $bx+c>0$ (if $b$ is also 0, only a constant is left, and the answer is either "all real numbers" or "no solution").

What is this calculation used for?

When a ball thrown upward is above a certain height (sports and physics)

The height of an object thrown straight up is a quadratic function of the time $t$ in seconds: $h = v_{0}t - 16t^{2}$ feet, where $v_{0}$ is the starting speed in feet per second. Asking "during what time is it at least a certain height?" gives a quadratic inequality directly.
For a ball thrown up at 64 feet per second, the time it is at least 48 feet high comes from $64t - 16t^{2} \ge 48$, that is, $t^{2} - 4t + 3 \le 0$, which gives 1 to 3 seconds after the throw, 2 seconds in all. Air resistance is ignored, but it gives a rough idea of how long fireworks stay in view or how long a fly ball is in the air.

Finding the top speed for stopping safely (road safety)

A car's stopping distance is the distance traveled while you react and hit the brakes (reaction distance) plus the distance traveled while braking (braking distance). The reaction distance grows in proportion to speed, but the braking distance grows with the square of speed, so the stopping distance is a quadratic function of speed.
With a 1-second reaction time and a road-tire friction coefficient of 0.7, the stopping distance is about $v + \dfrac{v^{2}}{45.08}$ feet ($v$ in feet per second). To stop within 200 feet of clear view, solve $v + \dfrac{v^{2}}{45.08} \le 200$: the speed must be at most about 75 feet per second, or about 51 mph. Speed limits and following-distance guidelines are based on quadratic relationships like this.

Finding the price range that makes a profit (business)

Raising the price increases the profit per item, but fewer items sell. When the number sold is a linear function of the price, the profit (profit per item × number sold − fixed costs) is a quadratic function of the price, and "the prices where the profit is 0 or more" come from a quadratic inequality.
Suppose an item costs $3 to make, $1000 - 200p$ items sell at a price of $p$ dollars, and fixed costs are $100. Solving $(p-3)(1000-200p) - 100 \ge 0$ shows a profit between about $3.30 and $4.70. The feeling that "too cheap or too expensive both lose money" is confirmed with numbers.

Getting enough area from a fixed length of material (design, DIY, farming)

For a rectangle with a fixed perimeter, choosing the length also fixes the width, so the area is a quadratic function of the length. "Which sizes give at least the area I need?" is a quadratic inequality.
Suppose you enclose a rectangular garden with 40 feet of fence and want at least 91 square feet. With a length of $x$ feet, the width is $20-x$ feet, so $x(20-x) \ge 91$, which gives a length from 7 to 13 feet. You can also see that the closer to a square, the larger the area (here the largest is 100 square feet at 10 feet).

Finding how wide an arch is tall enough (construction and transportation)

The cross section of a parabolic arch bridge or tunnel can be written as a quadratic function. "How far from the center can a tall vehicle pass?" is exactly a quadratic inequality.
Suppose an arch 80 feet wide and 16 feet high at the center is $y = -0.01x^{2} + 16$ ($x$ is the distance from the center). The part at least 15 feet high comes from $-0.01x^{2} + 16 \ge 15$: 10 feet on each side of the center, 20 feet wide in all. Decisions such as keeping large trucks near the center or posting a height limit are based on calculations like this.

Formulas and graphs

Positive $x^{2}$ coefficient and two roots ($>0$ is "outside the two roots")

Graph

Standard notation (the usual math form)

$x$ $<$ $\alpha$ $\ \text{or}\ $ $\beta$ $<$ $x$

In words (symbols replaced with words)

③ $x$: solution $<$ ① $\alpha$: smaller root $\ \text{or}\ $ ② $\beta$: larger root $<$ $x$: solution

The formula in words

① Let the two roots of the quadratic equation $ax^{2}+bx+c=0$ be $\alpha < \beta$. The part less than $\alpha$: smaller root and

② the part greater than $\beta$: larger root (that is, outside the two roots)

③ is $x$: solution (because a parabola that opens upward is above the $x$-axis outside the two crossing points)

Quick example

The solution of $x^{2}-3x+2>0$ (since $x^{2}-3x+2=(x-1)(x-2)$, the roots of $=0$ are $x=1,\ 2$) is

$x$: solution $<$ smaller root (1) $\ \text{or}\ $ larger root (2) $<$ $x$: solution

$(x-1)(x-2) > 0$

$x < 1 \ \text{or}\ 2 < x$

Key idea

Think of a quadratic inequality not as "a calculation" but as "reading off where the graph is above (or below) the $x$-axis". That cuts down a lot on what you have to memorize. A parabola with a positive $x^{2}$ coefficient opens upward (a valley shape). When it crosses the $x$-axis at two points, only the middle part, including the bottom of the valley, dips below the $x$-axis, and the parts outside rise above it. So the solution of $>0$ is outside the roots, and the solution of $<0$ is between them. "Outside" is two separate ranges, so write them joined by "or", as in $x<1$ or $2<x$. Writing them as one chain such as $2<x<1$ is wrong, because it would mean $x$ is greater than 2 and less than 1 at the same time. When the sign includes "equal to", as in $\ge 0$, the boundary values themselves are included: $x \le 1$ or $2 \le x$.

Positive $x^{2}$ coefficient and two roots ($<0$ is "between the two roots")

Graph

Standard notation (the usual math form)

$\alpha$ $<$ $x$ $<$ $\beta$

In words (symbols replaced with words)

① $\alpha$: smaller root $<$ ③ $x$: solution $<$ ② $\beta$: larger root

The formula in words

① The part greater than $\alpha$: smaller root and

② less than $\beta$: larger root (that is, between the two roots)

③ is $x$: solution (because a parabola that opens upward dips below the $x$-axis between the two crossing points)

Quick example

The solution of $x^{2}-3x+2<0$ (the roots of $=0$ are $x=1,\ 2$, as before) is

smaller root (1) $<$ $x$: solution $<$ larger root (2)

$(x-1)(x-2) < 0$

$1 < x < 2$

Key idea

With the same $x^{2}-3x+2$, just flipping the inequality sign swaps the solution from "outside" to "between". If the left side is the same, the boundary values (the roots of $=0$) are always the same; the only change is which side of the boundaries you take. $1<x<2$ means the $x$ values that satisfy both $1<x$ and $x<2$ at the same time: "greater than 1 and less than 2". Solving several inequalities at once like this is called a compound inequality, and the form $A<B<C$ writes it as one chain. When the sign includes "equal to", as in $\le 0$, it becomes $1 \le x \le 2$, and the endpoints 1 and 2 are also solutions.

The discriminant (decides the form of the solution)

Graph

Standard notation (the usual math form)

$D$ $=$ $b^{2}$ $-$ $4ac$

In words (symbols replaced with words)

③ $D$: discriminant $=$ ① $b^{2}$: the $x$ coefficient squared $-$ ② $4ac$: 4 times the $x^{2}$ coefficient times the constant

The formula in words

① From $b^{2}$: the $x$ coefficient squared

② subtract $4ac$: 4 times the $x^{2}$ coefficient $a$ times the constant $c$

③ and you get $D$: discriminant (its sign alone decides how the parabola meets the $x$-axis, which is the form of the solution)

Quick example

The discriminant of $x^{2}-3x+2$ is

$D$: discriminant $=$ $-3$ squared $-$ 4 times 1 times 2

$D = (-3)^{2} - 4 \times 1 \times 2 = 9 - 8 = 1 > 0$

Key idea

The discriminant tells you how many times the parabola meets the $x$-axis: at two points if $D>0$, touching at one point if $D=0$, and not at all if $D<0$. Even for a quadratic you cannot factor, this one value decides the form of the solution. When the $x^{2}$ coefficient is positive (the parabola opens upward), the four signs give these solutions: $D>0$ (crosses at two points): $>0$ and $\ge 0$ are outside the two roots; $<0$ and $\le 0$ are between them. $D=0$ (touches at one point): with the touching point at $x=\alpha$, $>0$ is $x \neq \alpha$ (only the touching point is left out), $\ge 0$ is all real numbers, $<0$ has no solution, and $\le 0$ is the single point $x=\alpha$. $D<0$ (never meets): the whole parabola is above the $x$-axis, so $>0$ and $\ge 0$ are all real numbers, and $<0$ and $\le 0$ have no solution. When $D=0$ or $D<0$, the answer is not a range, and these cases are often missed on tests. Sketch the graph and ask "does any part of the parabola dip below the $x$-axis?", and you will not get lost.

Negative $x^{2}$ coefficient (multiply both sides by $-1$ to make it positive)

Graph

Standard notation (the usual math form)

$ax^{2}+bx+c$ $< 0$

$\Longleftrightarrow$ $-ax^{2}-bx-c$ $> 0$

In words (symbols replaced with words)

① $ax^{2}+bx+c$: a quadratic with negative $a$ $< 0$

$\Longleftrightarrow$ ② $-ax^{2}-bx-c$: every sign changed (positive $x^{2}$ coefficient) $> 0$

The formula in words

① The inequality that $ax^{2}+bx+c$: a quadratic with a negative $x^{2}$ coefficient $a$ is less than 0

② can be rewritten as $-ax^{2}-bx-c$: the quadratic with every sign changed (positive $x^{2}$ coefficient) is greater than 0 (both sides were multiplied by the negative number $-1$, so the inequality sign flips too)

Quick example

Multiplying both sides of $-x^{2}+4x-3 \ge 0$ by $-1$ gives

$-x^{2}+4x-3$ ($x^{2}$ coefficient $-1$) $\ge 0$

$\Longleftrightarrow$ $x^{2}-4x+3$ ($x^{2}$ coefficient $1$) $\le 0$

$-x^{2}+4x-3 \ge 0 \Longleftrightarrow x^{2}-4x+3 \le 0$

$(x-1)(x-3) \le 0$

$1 \le x \le 3$

Key idea

If the $x^{2}$ coefficient stays negative, the parabola opens downward (a hill shape), and "outside" and "between" swap places, which is confusing. Multiply both sides by $-1$ first to turn it into an upward-opening parabola, and you only need to remember one pattern. In an inequality, multiplying or dividing both sides by a negative number flips the sign. For example, multiplying both sides of $2<3$ by $-1$ gives $-2>-3$, because changing signs flips left and right on the number line. Equations do not have this issue; it only matters for inequalities. Multiplying both sides by $-1$ does not change the discriminant $D$ ($(-b)^{2}-4(-a)(-c)=b^{2}-4ac$). In the example, $-x^{2}+4x-3$ and $x^{2}-4x+3$ both have $D=4$.

To solve a quadratic inequality, first make the $x^{2}$ coefficient positive (if it is negative, multiply both sides by $-1$ and flip the sign), then split into cases by the sign of the discriminant $D=b^{2}-4ac$. When $D>0$, $>0$ is outside the two roots and $<0$ is between them; when $D=0$ or $D<0$, the answer can be "all real numbers", "no solution" or "just one point". Picture an upward-opening parabola and read off where it is above the $x$-axis, and every case works the same way.

Symbols and terms

Symbols

$a,\ b,\ c$	a, b, c	The coefficients of the quadratic inequality $ax^{2}+bx+c>0$: $a$ is the coefficient of $x^{2}$, $b$ the coefficient of $x$, and $c$ the constant term. By custom, letters near the start of the alphabet, $a,\ b,\ c$, stand for numbers that are given in advance.
$x$	ex	The unknown number. A quadratic inequality asks "from where to where do the values of $x$ make it true?", so the answer is a range, not a single number. By custom, letters near the end of the alphabet, $x,\ y,\ z$, are used for unknowns.
$D$	dee	The discriminant, $b^{2}-4ac$. $D$ is the first letter of "discriminant" (the thing that tells cases apart). Its sign alone decides how the parabola meets the $x$-axis (the form of the solution).
$\alpha,\ \beta$	alpha, beta	Names for the two roots of the quadratic equation $ax^{2}+bx+c=0$. They are the first and second letters of the Greek alphabet, and by custom $\alpha < \beta$ ($\alpha$ is the smaller one). In a quadratic inequality, these two are the boundaries of the solution set.
$>,\ <$	greater than, less than	Inequality signs. $A>B$ means "$A$ is greater than $B$", and $A<B$ means "$A$ is less than $B$"; the open side always faces the larger number. They do not include "equal to", so the boundary values themselves are not solutions.
$\ge,\ \le$	greater than or equal to, less than or equal to	Inequality signs that include "equal to". $A \ge B$ means "$A$ is greater than or equal to $B$", and $A \le B$ means "$A$ is less than or equal to $B$". The boundary values themselves are solutions too. Some countries, such as Japan, write them with a double line under the sign (≧, ≦); the meaning is the same.
$\neq$	not equal to	The symbol for "is not equal to". $x \neq 3$ means "$x$ can be any value except 3". In quadratic inequalities, this form of answer appears when the parabola touches the $x$-axis ($D=0$).
$\sqrt{\phantom{x}}$	square root	The symbol for the positive number whose square is the number inside (the square root). $\sqrt{5}$ is "the positive number whose square is 5", about 2.236. It is said to come from a stylized letter r, for the Latin word radix, "root". It appears when the boundary values do not come out exactly.
$\Longleftrightarrow$	if and only if (is equivalent to)	A symbol that says the left and right sides mean the same thing. $P \Longleftrightarrow Q$ means "when $P$ is true, $Q$ is true, and the other way around". When you rewrite an inequality, take care to keep this relation (so that the solution does not change).
$y$	why	The value of the quadratic function $y=ax^{2}+bx+c$, the "height" of the parabola. Solving $ax^{2}+bx+c>0$ is the same as finding the range of $x$ where this height is greater than 0 (where the graph is above the $x$-axis).

Terms

quadratic inequality	An inequality with an $x^{2}$ term, such as $ax^{2}+bx+c>0$. The answer is a range of $x$, not a single number. Taught in Algebra 1 and Algebra 2.
inequality	A statement that compares two amounts with $>$, $<$, $\ge$ or $\le$. Adding or subtracting the same number on both sides does not change the direction, but multiplying or dividing by a negative number flips it.
quadratic function	A function of the form $y=ax^{2}+bx+c$ ($a \neq 0$). Its graph is a parabola. A quadratic inequality can be restated as reading off where this graph is above or below the $x$-axis.
parabola	The smooth, symmetric curve that is the graph of a quadratic function. A ball thrown at an angle follows this shape.
opens upward	A parabola in a valley shape (both ends pointing up). This happens when the $x^{2}$ coefficient is positive. When the coefficient is negative, it is a hill shape and "opens downward". An upward-opening parabola is also called concave up.
discriminant	$D=b^{2}-4ac$. It tells how many real solutions the quadratic equation has (how many times the parabola meets the $x$-axis): 2 if $D>0$, 1 (a double root) if $D=0$, and 0 if $D<0$.
real solution	A solution that is an ordinary number on the number line (a real number). When $D<0$, there are no real solutions, and the parabola does not meet the $x$-axis.
double root	The solution when $D=0$. Two equal solutions come together, as in $(x-3)^{2}=0$, and the parabola touches the $x$-axis at one point.
quadratic formula	$x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$. It solves quadratic equations even when you cannot factor them, and in quadratic inequalities it gives the boundary values.
factoring	Rewriting an expression as a product, as in $x^{2}-3x+2=(x-1)(x-2)$. In a quadratic inequality, if you can factor, you see the boundary values at a glance.
boundary value	A value where the solution set begins or ends. In a quadratic inequality, these are the solutions of $ax^{2}+bx+c=0$, the $x$-coordinates where the parabola meets the $x$-axis. They are included with $\ge$ or $\le$ and left out with $>$ or $<$. They are also called critical values.
solution set	All the values of $x$ that make the inequality true. It can be one range, such as $1<x<2$, or two separate ranges, such as $x<1$ or $2<x$.
compound inequality	Finding the $x$ that satisfies several inequalities at once. The range that satisfies both $1<x$ and $x<2$ is written as one chain, $1<x<2$.
no solution	There is no $x$ at all that makes the inequality true. It happens, for example, with $x^{2}+2x+5<0$, where an upward-opening parabola never dips below the $x$-axis.
all real numbers	The inequality is true for any $x$. With a positive $x^{2}$ coefficient, this is the answer for $>0$ or $\ge 0$ when $D<0$ (the parabola never meets the $x$-axis). When $D=0$, $\ge 0$ also gives all real numbers, because the touching point counts through "equal to". (With the same $D<0$, $<0$ or $\le 0$ has no solution.)
number line	A picture that places numbers on a straight line. To show the solution of a quadratic inequality, draw the range as a thick line, with a closed circle at an included boundary and an open circle at one that is not included.
moving a term	Moving a term of an inequality or equation to the other side and changing its sign. If there is an expression on the right side, move it to the left so that the right side is 0 before solving.
constant term	A number-only term with no $x$, such as the 5 in $x^{2}+3x+5$. It is called constant because it does not change whatever value $x$ has.
coefficient	The number in front of a letter. In $3x^{2}$, the coefficient is 3. When no number is written, as in $x^{2}$, the coefficient is 1.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back to review these topics is the fastest way forward.

Properties of inequalities (Grade 7 and Algebra 1)	Knowing the difference between the four signs $>$, $<$, $\ge$ and $\le$ ("or equal to" includes the boundary value; "greater than" and "less than" do not) Knowing that adding or subtracting the same number on both sides keeps the sign, but multiplying or dividing by a negative number flips it (for example, multiplying both sides of $2<3$ by $-1$ gives $-2>-3$) Knowing that a chain such as $1<x<2$ combines two conditions, "greater than 1 and less than 2"
Factoring (Algebra 1)	Being able to factor by finding two numbers that add to $-3$ and multiply to $2$, as in $x^{2}-3x+2=(x-1)(x-2)$ Being able to factor when the $x^{2}$ coefficient is not 1, as in $2x^{2}-5x+2=(2x-1)(x-2)$ Being able to expand a square such as $(x-3)^{2}$ into $x^{2}-6x+9$
Solving quadratic equations (Algebra 1)	Being able to solve by factoring, using the fact that if $AB=0$, then $A=0$ or $B=0$ (the zero product property) Being able to solve quadratic equations you cannot factor with the quadratic formula $x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ Being able to simplify a radical by taking square factors out, as in $\sqrt{20}=2\sqrt{5}$
Graphs of quadratic functions (Algebra 1)	Knowing that the graph of $y=ax^{2}+bx+c$ is a parabola that opens upward (a valley) when $a$ is positive and opens downward (a hill) when $a$ is negative Knowing that the $x$-coordinates where the graph meets the $x$-axis are the solutions of $ax^{2}+bx+c=0$ Being able to read that where the height $y$ is positive the graph is above the $x$-axis, and where it is negative the graph is below
Square roots (Grade 8)	Knowing that $\sqrt{5}$ is "the positive number whose square is 5", about 2.236 Knowing that the square root of a perfect square is a whole number, as in $\sqrt{9}=3$
The number line and ranges (Grades 6–7)	Knowing that numbers get larger to the right on the number line ($-2$ is greater than $-5$) Knowing the convention of drawing a range as a thick line on the number line, with a closed circle at an included endpoint and an open circle at one that is not included

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table for two roots, outside (ax²+bx+c > 0)

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Smaller root α	=(-B2-SQRT(B4))/(2*B1)
Larger root β	=(-B2+SQRT(B4))/(2*B1)
Solution set (outside)	="x < "&B5&" or "&B6&" < x"

Table for two roots, between (ax²+bx+c < 0)

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Smaller root α	=(-B2-SQRT(B4))/(2*B1)
Larger root β	=(-B2+SQRT(B4))/(2*B1)
Solution set (between)	=B5&" < x < "&B6

Table to find the discriminant D = b²−4ac

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Number of x-axis crossings	=IF(B4>0,2,IF(B4=0,1,0))

Table to check the sign change when the x² coefficient is negative

Coefficient of x², a	-1
Coefficient of x, b	4
Constant term c	-3
x² coefficient times −1	=-B1
x coefficient times −1	=-B2
Constant term times −1	=-B3
Discriminant after the change (same as before)	=B5^2-4B4B6

After pasting, the top three rows (the coefficients a, b and c) are your inputs and the lower rows are calculated automatically.
"^" means a power, "SQRT" is the square root, and "&" joins text together.
The first and second tables both use a=1, b=−3, c=2 (that is, x²−3x+2). The discriminant is 1, the boundary values are 1 and 2, and the solutions are "x < 1 or 2 < x" (outside) and "1 < x < 2" (between). If you enter coefficients that make the discriminant negative, SQRT returns an error, so check the number of crossings with the third table first.
The third table finds only the discriminant, and the IF function also decides the number of x-axis crossings (2, 1 or 0).
The fourth table changes every sign when the x² coefficient is negative (a=−1, b=4, c=−3). The new coefficients are 1, −4 and 3, and you can check that the discriminant is 4 both before and after the change.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table for two roots, outside (ax²+bx+c > 0)

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Smaller root α	=(-B2-SQRT(B4))/(2*B1)
Larger root β	=(-B2+SQRT(B4))/(2*B1)
Solution set (outside)	="x < "&B5&" or "&B6&" < x"

Table for two roots, between (ax²+bx+c < 0)

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Smaller root α	=(-B2-SQRT(B4))/(2*B1)
Larger root β	=(-B2+SQRT(B4))/(2*B1)
Solution set (between)	=B5&" < x < "&B6

Table to find the discriminant D = b²−4ac

Coefficient of x², a	1
Coefficient of x, b	-3
Constant term c	2
Discriminant D = b^2-4ac	=B2^2-4B1B3
Number of x-axis crossings	=IF(B4>0,2,IF(B4=0,1,0))

Table to check the sign change when the x² coefficient is negative

Coefficient of x², a	-1
Coefficient of x, b	4
Constant term c	-3
x² coefficient times −1	=-B1
x coefficient times −1	=-B2
Constant term times −1	=-B3
Discriminant after the change (same as before)	=B5^2-4B4B6

The same formulas as in Excel work as is (SQRT, IF and & have the same names). Copy the whole table, paste it into cell A1, and replace the coefficients with your own numbers.

How to calculate it in Python

from fractions import Fraction
import math

# Coefficients of the quadratic inequality (the left side a*x^2 + b*x + c is compared with 0) and the sign (">", ">=", "<" or "<=")
a = Fraction(1)
b = Fraction(-3)
c = Fraction(2)
sign = ">"

# Make the x^2 coefficient positive (if negative, multiply both sides by -1 and flip the sign)
if a < 0:
    a, b, c = -a, -b, -c
    sign = {">": "<", ">=": "<=", "<": ">", "<=": ">="}[sign]

discriminant = b * b - 4 * a * c
is_greater = sign in (">", ">=")
has_equal = sign in (">=", "<=")

if discriminant > 0:
    root = math.sqrt(discriminant)
    alpha = (-b - root) / (2 * a)
    beta = (-b + root) / (2 * a)
    mark = "<=" if has_equal else "<"
    if is_greater:
        answer = f"x {mark} {alpha} or {beta} {mark} x"   # outside the two roots
    else:
        answer = f"{alpha} {mark} x {mark} {beta}"        # between the two roots
elif discriminant == 0:
    alpha = -b / (2 * a)
    answer = {">": f"x != {alpha}", ">=": "all real numbers",
              "<": "no solution", "<=": f"x = {alpha}"}[sign]
else:
    answer = "all real numbers" if is_greater else "no solution"

print(f"Discriminant D = {discriminant}")
print(f"Solution set: {answer}")

Runs with the standard library only. If you make the coefficients fractions.Fraction, the sign of the discriminant is not affected by decimal rounding errors. This example is x² − 3x + 2 > 0, and running it prints two lines, "Discriminant D = 1" and "Solution set: x < 1.0 or 2.0 < x". Change the coefficients and the sign and run it.

How to write it in LaTeX and other math languages (copy and paste)

Positive $x^{2}$ coefficient and two roots ($>0$ is "outside the two roots")

x < α or β < x

x < \alpha \ \text{or}\ \beta < x

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>x</mi><mo>&lt;</mo><mi>&#x3B1;</mi>
    <mtext>&#xA0;or&#xA0;</mtext>
    <mi>&#x3B2;</mi><mo>&lt;</mo><mi>x</mi>
  </mrow>
</math>

x < alpha " or " beta < x

Reduce[a x^2 + b x + c > 0 && a > 0, x, Reals]

solve(a*x^2 + b*x + c > 0, x);

solve(a*x^2 + b*x + c > 0, x)

x < α or β < x

Positive $x^{2}$ coefficient and two roots ($<0$ is "between the two roots")

α < x < β

\alpha < x < \beta

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>&#x3B1;</mi><mo>&lt;</mo><mi>x</mi><mo>&lt;</mo><mi>&#x3B2;</mi>
  </mrow>
</math>

alpha < x < beta

Reduce[a x^2 + b x + c < 0 && a > 0, x, Reals]

solve(a*x^2 + b*x + c < 0, x);

solve(a*x^2 + b*x + c < 0, x)

α < x < β

The discriminant (decides the form of the solution)

D = b² − 4ac

D = b^{2} - 4ac

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>D</mi><mo>=</mo>
    <msup><mi>b</mi><mn>2</mn></msup>
    <mo>&#x2212;</mo>
    <mn>4</mn><mi>a</mi><mi>c</mi>
  </mrow>
</math>

D = b^2 - 4ac

Discriminant[a x^2 + b x + c, x]

discrim(a*x^2 + b*x + c, x);

D = b^2 - 4*a*c;

D = b^2 - 4ac

Negative $x^{2}$ coefficient (multiply both sides by $-1$ to make it positive)

ax² + bx + c < 0 ⟺ −ax² − bx − c > 0

ax^{2} + bx + c < 0 \Longleftrightarrow -ax^{2} - bx - c > 0

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>a</mi><msup><mi>x</mi><mn>2</mn></msup>
    <mo>+</mo><mi>b</mi><mi>x</mi><mo>+</mo><mi>c</mi>
    <mo>&lt;</mo><mn>0</mn>
    <mo>&#x27FA;</mo>
    <mo>&#x2212;</mo><mi>a</mi><msup><mi>x</mi><mn>2</mn></msup>
    <mo>&#x2212;</mo><mi>b</mi><mi>x</mi><mo>&#x2212;</mo><mi>c</mi>
    <mo>&gt;</mo><mn>0</mn>
  </mrow>
</math>

ax^2 + bx + c < 0 hArr -ax^2 - bx - c > 0

Reduce[-a x^2 - b x - c > 0, x, Reals]

solve(-a*x^2 - b*x - c > 0, x);

solve(-a*x^2 - b*x - c > 0, x)

ax^2 + bx + c < 0 ⟺ -ax^2 - bx - c > 0

How to have ChatGPT do the calculation

You are a calculation assistant for algebra. Solve the following problem by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

Solve the quadratic inequality x² − 3x + 2 > 0.
Show each of the following:
1. The value of the discriminant D = b² − 4ac
2. The boundary values (the solutions of x² − 3x + 2 = 0)
3. The solution set (also say whether it is "outside the two roots", "between the two roots", "all real numbers" or "no solution")
4. If the x² coefficient were negative, also show the step of multiplying both sides by −1 and flipping the sign

In Python, use both sympy's solve_univariate_inequality and an exact discriminant calculation with fractions, and answer only after confirming that the results agree.

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\(a,\ b,\ c\)	a, b, c	The coefficients of the quadratic inequality \(ax^{2}+bx+c>0\): \(a\) is the coefficient of \(x^{2}\), \(b\) the coefficient of \(x\), and \(c\) the constant term. By custom, letters near the start of the alphabet, \(a,\ b,\ c\), stand for numbers that are given in advance.
\(x\)	ex	The unknown number. A quadratic inequality asks "from where to where do the values of \(x\) make it true?", so the answer is a range, not a single number. By custom, letters near the end of the alphabet, \(x,\ y,\ z\), are used for unknowns.
\(D\)	dee	The discriminant, \(b^{2}-4ac\). \(D\) is the first letter of "discriminant" (the thing that tells cases apart). Its sign alone decides how the parabola meets the \(x\)-axis (the form of the solution).
\(\alpha,\ \beta\)	alpha, beta	Names for the two roots of the quadratic equation \(ax^{2}+bx+c=0\). They are the first and second letters of the Greek alphabet, and by custom \(\alpha < \beta\) (\(\alpha\) is the smaller one). In a quadratic inequality, these two are the boundaries of the solution set.
\(>,\ <\)	greater than, less than	Inequality signs. \(A>B\) means "\(A\) is greater than \(B\)", and \(A<B\) means "\(A\) is less than \(B\)"; the open side always faces the larger number. They do not include "equal to", so the boundary values themselves are not solutions.
\(\ge,\ \le\)	greater than or equal to, less than or equal to	Inequality signs that include "equal to". \(A \ge B\) means "\(A\) is greater than or equal to \(B\)", and \(A \le B\) means "\(A\) is less than or equal to \(B\)". The boundary values themselves are solutions too. Some countries, such as Japan, write them with a double line under the sign (≧, ≦); the meaning is the same.
\(\neq\)	not equal to	The symbol for "is not equal to". \(x \neq 3\) means "\(x\) can be any value except 3". In quadratic inequalities, this form of answer appears when the parabola touches the \(x\)-axis (\(D=0\)).
\(\sqrt{\phantom{x}}\)	square root	The symbol for the positive number whose square is the number inside (the square root). \(\sqrt{5}\) is "the positive number whose square is 5", about 2.236. It is said to come from a stylized letter r, for the Latin word radix, "root". It appears when the boundary values do not come out exactly.
\(\Longleftrightarrow\)	if and only if (is equivalent to)	A symbol that says the left and right sides mean the same thing. \(P \Longleftrightarrow Q\) means "when \(P\) is true, \(Q\) is true, and the other way around". When you rewrite an inequality, take care to keep this relation (so that the solution does not change).
\(y\)	why	The value of the quadratic function \(y=ax^{2}+bx+c\), the "height" of the parabola. Solving \(ax^{2}+bx+c>0\) is the same as finding the range of \(x\) where this height is greater than 0 (where the graph is above the \(x\)-axis).

quadratic inequality	An inequality with an \(x^{2}\) term, such as \(ax^{2}+bx+c>0\). The answer is a range of \(x\), not a single number. Taught in Algebra 1 and Algebra 2.
inequality	A statement that compares two amounts with \(>\), \(<\), \(\ge\) or \(\le\). Adding or subtracting the same number on both sides does not change the direction, but multiplying or dividing by a negative number flips it.
quadratic function	A function of the form \(y=ax^{2}+bx+c\) (\(a \neq 0\)). Its graph is a parabola. A quadratic inequality can be restated as reading off where this graph is above or below the \(x\)-axis.
parabola	The smooth, symmetric curve that is the graph of a quadratic function. A ball thrown at an angle follows this shape.
opens upward	A parabola in a valley shape (both ends pointing up). This happens when the \(x^{2}\) coefficient is positive. When the coefficient is negative, it is a hill shape and "opens downward". An upward-opening parabola is also called concave up.
discriminant	\(D=b^{2}-4ac\). It tells how many real solutions the quadratic equation has (how many times the parabola meets the \(x\)-axis): 2 if \(D>0\), 1 (a double root) if \(D=0\), and 0 if \(D<0\).
real solution	A solution that is an ordinary number on the number line (a real number). When \(D<0\), there are no real solutions, and the parabola does not meet the \(x\)-axis.
double root	The solution when \(D=0\). Two equal solutions come together, as in \((x-3)^{2}=0\), and the parabola touches the \(x\)-axis at one point.
quadratic formula	\(x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). It solves quadratic equations even when you cannot factor them, and in quadratic inequalities it gives the boundary values.
factoring	Rewriting an expression as a product, as in \(x^{2}-3x+2=(x-1)(x-2)\). In a quadratic inequality, if you can factor, you see the boundary values at a glance.
boundary value	A value where the solution set begins or ends. In a quadratic inequality, these are the solutions of \(ax^{2}+bx+c=0\), the \(x\)-coordinates where the parabola meets the \(x\)-axis. They are included with \(\ge\) or \(\le\) and left out with \(>\) or \(<\). They are also called critical values.
solution set	All the values of \(x\) that make the inequality true. It can be one range, such as \(1<x<2\), or two separate ranges, such as \(x<1\) or \(2<x\).
compound inequality	Finding the \(x\) that satisfies several inequalities at once. The range that satisfies both \(1<x\) and \(x<2\) is written as one chain, \(1<x<2\).
no solution	There is no \(x\) at all that makes the inequality true. It happens, for example, with \(x^{2}+2x+5<0\), where an upward-opening parabola never dips below the \(x\)-axis.
all real numbers	The inequality is true for any \(x\). With a positive \(x^{2}\) coefficient, this is the answer for \(>0\) or \(\ge 0\) when \(D<0\) (the parabola never meets the \(x\)-axis). When \(D=0\), \(\ge 0\) also gives all real numbers, because the touching point counts through "equal to". (With the same \(D<0\), \(<0\) or \(\le 0\) has no solution.)
number line	A picture that places numbers on a straight line. To show the solution of a quadratic inequality, draw the range as a thick line, with a closed circle at an included boundary and an open circle at one that is not included.
moving a term	Moving a term of an inequality or equation to the other side and changing its sign. If there is an expression on the right side, move it to the left so that the right side is 0 before solving.
constant term	A number-only term with no \(x\), such as the 5 in \(x^{2}+3x+5\). It is called constant because it does not change whatever value \(x\) has.
coefficient	The number in front of a letter. In \(3x^{2}\), the coefficient is 3. When no number is written, as in \(x^{2}\), the coefficient is 1.

Quadratic Inequality Calculator (Solution Set, Discriminant Cases and Graph)

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How to calculate it in Excel

How to calculate it in Google Sheets

How to calculate it in Python

How to write it in LaTeX and other math languages (copy and paste)

How to have ChatGPT do the calculation

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