Quadratic Formula Calculator (Discriminant and Complex Roots)

Enter the coefficients a, b and c of the quadratic equation ax² + bx + c = 0. The equation below is linked to the input fields, so you can also edit the coefficients directly in it. The quadratic formula gives the solutions, along with the discriminant, the type of solutions and the steps.

a (number in front of x²)

b (number in front of x)

c (constant term)

Decimals, negative numbers and fractions such as 3/4 can be used. If a term is missing, as in x² − 4 = 0, enter 0 for its coefficient.

Result and graph

Enter the coefficients a, b and c in the fields on the left and press "Calculate". The result will appear here.

[ CLOSE ]

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

Enter the coefficients $a, b, c$ and the quadratic formula solves $ax^2 + bx + c = 0$ on the spot
It also shows the value of the discriminant $D = b^2 - 4ac$ and the type of solutions (two different real solutions, a double root, or two complex solutions)
When the solutions are not real numbers, it still finds the two complex solutions, such as $-0.5 \pm 0.866i$ (a conjugate pair that differs only in the sign of the imaginary part)
Coefficients can be decimals, negative numbers or fractions such as 3/4
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

When a = 0, the equation is not quadratic, so it is solved as the linear equation $bx + c = 0$ (the solution is $x = -c \div b$). a and b cannot both be 0.

What is this calculation used for?

Height and time of a ball thrown upward or a firework (physics and sports)

If you throw a ball straight up at 64 feet per second, its height after t seconds is about 64t − 16t² feet (using gravity of about 32 ft/s² and ignoring air resistance). "When is it 48 feet high?" means solving the quadratic equation −16t² + 64t − 48 = 0, and the quadratic formula gives t = 1 second and 3 seconds (once on the way up and once on the way down).
From planning the height where fireworks burst to the path of a basketball shot, the motion of anything thrown is tied to quadratic equations.

Finding the size of a garden bed or a lot from its area (DIY and real estate)

"I want a rectangular garden bed with an area of exactly 28 square feet, 3 feet longer than it is wide." With width x, you get x(x + 3) = 28, or x² + 3x − 28 = 0. The quadratic formula gives x = 4 and x = −7, but a length cannot be negative, so the bed is 4 feet wide and 7 feet long.
In real life, using a quadratic equation includes this last step: choosing which of the two solutions makes sense (checking whether each answer makes sense).

Working backward to a compound interest rate (money)

What annual rate turns $10,000 into $12,100 in 2 years? With the rate r, 10000(1 + r)² = 12100, which simplifies to r² + 2r − 0.21 = 0. The quadratic formula gives r = 0.1 and r = −2.1, and the one that makes sense as a rate, r = 0.1 (10% per year), is the answer.
Money calculations that "multiply by the same factor twice", like compound interest, lead to quadratic equations, so you can work back from a target amount to the return you need.

Designing bridges and arches (architecture and civil engineering)

The shape of an arch bridge or the cable of a suspension bridge is usually designed with a parabola, y = ax² + bx + c. Finding "how far from the center the height above the ground is a certain value" is exactly solving a quadratic equation.
When designers and civil engineers set the shape of a structure with formulas, the quadratic formula is one of the basics.

Estimating the prices where profit is zero (business)

"Raising the price increases the profit per item, but fewer items sell." In a simple model of this, profit is a quadratic function of the price. The prices where the profit is exactly 0 are the solutions of that quadratic equation, and they give a rough upper and lower limit for pricing.
Real markets are more complicated, but a first estimate of "where the profit hill ends" with a quadratic equation is a common starting point in business analysis.

Formulas and graphs

The quadratic formula

Graph

Standard notation (the usual math form)

$x$ $=$ $($ $-b$ $\pm$ $\sqrt{b^2 - 4ac}$ $)$ $\div$ $($ $2a$ $)$

In words (symbols replaced with words)

④ $x$: solution $=$ $($ ① $-b$: opposite of the $x$ coefficient $\pm$ ② $\sqrt{b^2 - 4ac}$: square root of the discriminant $)$ $\div$ $($ ③ $2a$: twice the $x^2$ coefficient $)$

The formula in words

① To $-b$: opposite of the $x$ coefficient

② add and subtract $\sqrt{b^2 - 4ac}$: square root of the discriminant (the two ways of ±),

③ then divide each by $2a$: twice the $x^2$ coefficient

④ and you get $x$: solution (because of ±, there are usually two solutions)

Quick example

The solutions of the quadratic equation x² + 3x − 4 = 0 (a = 1, b = 3, c = −4) are

$x$: solution $=$ $($ opposite of $b$ (−3) $\pm$ square root of the discriminant 25 (5) $)$ $\div$ $($ twice $a$ (2) $)$

$x = \dfrac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-4)}}{2 \times 1} = \dfrac{-3 \pm \sqrt{25}}{2} = \dfrac{-3 \pm 5}{2}$

$x = \dfrac{-3 + 5}{2} = 1, \quad x = \dfrac{-3 - 5}{2} = -4$

Key idea

The quadratic formula is the "answer formula" you get by solving $ax^2 + bx + c = 0$ ($a \neq 0$) all the way through by completing the square. Whatever the coefficients, you just put in the values of $a, b, c$ to get the solutions. ± (plus or minus) means that "adding and subtracting both give a solution", so there are usually two solutions. Equations that can be solved by factoring can of course be solved this way too, and the answers always match.

The discriminant (tells you the type of solutions)

Graph

Standard notation (the usual math form)

$D$ $=$ $b$ $2$ $-$ $4$ $\times$ $a$ $\times$ $c$

In words (symbols replaced with words)

⑥ $D$: discriminant $=$ ① $b$: $x$ coefficient ② squared $-$ ③ the number $4$ $\times$ ④ $a$: $x^2$ coefficient $\times$ ⑤ $c$: constant term

The formula in words

① Take $b$: $x$ coefficient

② and square it. From that, subtract the product of

③ the number $4$

④ $a$: $x^2$ coefficient and

⑤ $c$: constant term

⑥ and you get $D$: discriminant

Quick example

The discriminant of x² + 3x − 4 = 0 (a = 1, b = 3, c = −4) is

$D$: discriminant $=$ $b$ (3) squared $-$ 4 $\times$ $a$ (1) $\times$ $c$ (−4)

$D = 3^2 - 4 \times 1 \times (-4) = 9 + 16 = 25$

Key idea

The discriminant $D$ is exactly what is under the √ in the quadratic formula. It tells you what type of solutions to expect before you calculate them. - $D > 0$: the square root is a positive real number, so there are two different real solutions (on the graph, the parabola crosses the x-axis at two points) - $D = 0$: the square root is 0, so there is a double root, exactly one real solution (the parabola touches the x-axis) - $D < 0$: the number under the √ is negative, so there is no real solution. If we extend numbers to include imaginary numbers, there are two complex solutions (the parabola does not meet the x-axis)

Double root (when the discriminant $D = 0$)

Graph

Standard notation (the usual math form)

$x$ $=$ $-b$ $\div$ $($ $2a$ $)$

In words (symbols replaced with words)

③ $x$: double root $=$ ① $-b$: opposite of the $x$ coefficient $\div$ $($ ② $2a$: twice the $x^2$ coefficient $)$

The formula in words

① Divide $-b$: opposite of the $x$ coefficient

② by $2a$: twice the $x^2$ coefficient

③ and you get $x$: double root (this is the only solution)

Quick example

The solution of x² + 2x + 1 = 0 (a = 1, b = 2, c = 1; discriminant D = 4 − 4 = 0) is

$x$: double root $=$ opposite of $b$ (−2) $\div$ $($ twice $a$ (2) $)$

$D = 2^2 - 4 \times 1 \times 1 = 0$

$x = \dfrac{-2}{2 \times 1} = -1$

Key idea

When $D = 0$, $\sqrt{0} = 0$, so the "± √D" part of the quadratic formula disappears and the two solutions become one. This is called a double root (or repeated root). It happens when the left side factors into a perfect square, as in $x^2 + 2x + 1 = (x + 1)^2$.

Two complex solutions (when the discriminant $D < 0$)

Graph

Standard notation (the usual math form)

$x$ $=$ $($ $-b$ $\pm$ $\sqrt{4ac - b^2}$ $\,i$ $)$ $\div$ $($ $2a$ $)$

In words (symbols replaced with words)

⑤ $x$: two complex solutions $=$ $($ ① $-b$: opposite of the $x$ coefficient $\pm$ ② $\sqrt{4ac - b^2}$: square root of $4ac - b^2$ ③ $\,i$ $)$ $\div$ $($ ④ $2a$: twice the $x^2$ coefficient $)$

The formula in words

① To $-b$: opposite of the $x$ coefficient

② add and subtract $\sqrt{4ac - b^2}$: square root of the positive number $4ac - b^2$ (the discriminant with its sign flipped)

③ times the imaginary unit $i$ (the number whose square is −1), the two ways of ±,

④ then divide each by $2a$: twice the $x^2$ coefficient

⑤ and you get $x$: two complex solutions (a conjugate pair that differs only in the sign of the imaginary part)

Quick example

The solutions of x² + x + 1 = 0 (a = 1, b = 1, c = 1; discriminant D = 1 − 4 = −3) are

$x$: two complex solutions $=$ $($ opposite of $b$ (−1) $\pm$ square root of $4ac - b^2 = 3$ $\,i$ $)$ $\div$ $($ twice $a$ (2) $)$

$x = \dfrac{-1 \pm \sqrt{3}\,i}{2} \approx -0.5 \pm 0.866\,i$

Key idea

When $D < 0$, the number under the √ is negative. No real number gives a negative number when squared, so there is no solution among the real numbers. So we bring in the imaginary unit $i$, "the number whose square is −1". Then we can write $\sqrt{-3} = \sqrt{3}\,i$, and the solutions can be written as two complex numbers. $4ac - b^2$ is the discriminant $b^2 - 4ac$ with its sign flipped, and it is always positive when $D < 0$, so its square root is safe to take.

When $a = 0$ (it becomes a linear equation)

Graph

Standard notation (the usual math form)

$x$ $=$ $-c$ $\div$ $b$

In words (symbols replaced with words)

③ $x$: solution of the linear equation $=$ ① $-c$: opposite of the constant term $\div$ ② $b$: $x$ coefficient

The formula in words

① Divide $-c$: opposite of the constant term

② by $b$: $x$ coefficient

③ and you get $x$: solution of the linear equation

Quick example

With a = 0, the solution of 2x + 4 = 0 (b = 2, c = 4) is

$x$: solution of the linear equation $=$ opposite of $c$ (−4) $\div$ $b$ (2)

$x = \dfrac{-4}{2} = -2$

Key idea

When $a = 0$, the $x^2$ term disappears and the equation becomes the linear equation $bx + c = 0$, not a quadratic. The quadratic formula has $2a$ in the denominator, so it assumes $a \neq 0$. This calculator does not treat a = 0 as an error; it solves the linear equation and returns one solution (only when a and b are both 0 is there no equation to solve, and you get an error).

The solutions of the quadratic equation ax² + bx + c = 0 come from putting the coefficients into the quadratic formula, x = (−b ± √(b² − 4ac)) ÷ (2a). The sign of the discriminant D = b² − 4ac, the part under the √, tells you before you calculate whether there are two solutions, one (a double root), or complex solutions.

Symbols and terms

Symbols

$a, b, c$	a, b, c	The coefficients of the quadratic equation. $a$ is the number in front of $x^2$, $b$ is the number in front of $x$, and $c$ is the constant term (the number-only term).
$x$	ex	The unknown number you want to find as the solution of the equation. A quadratic equation has at most two solutions.
$D$	dee (discriminant)	The discriminant, $D = b^2 - 4ac$, from the first letter of "discriminant". Some books write it with the Greek letter $\Delta$ (delta).
$\sqrt{\phantom{9}}$	square root	The symbol for "the number, 0 or greater, whose square is the number inside". $\sqrt{25} = 5$ and $\sqrt{2} \approx 1.414$.
$\pm$	plus or minus	A symbol that writes "both the + case and the − case" at once. In the quadratic formula, it means that both the sum and the difference are solutions.
$i$	i (imaginary unit)	The number whose square is $-1$ ($i^2 = -1$). It is used to write the solutions when the number under the √ is negative.
$x_1, x_2$	x sub 1, x sub 2	Names for the two solutions. On this page, the + side of ± is $x_1$ and the − side is $x_2$.

Terms

quadratic equation	An equation with an $x^2$ (squared) term. It can be written in the form $ax^2 + bx + c = 0$ ($a \neq 0$), called standard form. Taught in Algebra 1.
coefficient	The number in front of a letter. In $3x$, the coefficient is 3. When reading coefficients, watch for hidden 1s and signs: $x^2 - x = 0$ means a = 1, b = −1, c = 0.
solution	A value of the letter that makes the equation true. To "solve" an equation means to find all of its solutions. The solutions of a quadratic are also called its roots.
quadratic formula	A formula that calculates the solutions of a quadratic equation directly from the coefficients a, b and c. Its strength is that it always works, even when you cannot see how to factor.
discriminant	The part under the √ in the quadratic formula, $b^2 - 4ac$. Its sign alone tells you the type of solutions (two real solutions, a double root, or two complex solutions).
double root	A solution where the two solutions come together as one. It happens when the discriminant is 0, and on the graph the parabola touches the x-axis. It is also called a repeated root.
real solution	A solution that is a real number (a number on the ordinary number line). There are real solutions when the discriminant is 0 or greater.
complex solution	A solution that is not a real number and is written with the imaginary unit $i$. When the discriminant is negative, the solutions are a pair of the form "real part ± imaginary part $i$", such as $-0.5 \pm 0.866i$ (complex conjugates). Also called imaginary solutions. Taught in Algebra 2.
completing the square	Rewriting an equation in the form $(x + p)^2 = q$ (a square equals a number). Doing this all the way with the letters a, b and c gives the quadratic formula.
factoring	Writing an expression as a product. If you can factor $x^2 + 3x - 4 = (x - 1)(x + 4)$, you get the solutions x = 1, −4 without the quadratic formula.
parabola	The shape of the graph of $y = ax^2 + bx + c$. The real solutions of the quadratic equation are the x-coordinates of the points where this parabola meets the x-axis (the x-intercepts).

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.

Arithmetic with negative numbers (Grade 7)	Knowing that squaring a negative number gives a positive number, as in $(-5)^2 = 25$ Being able to multiply while keeping track of signs, as in $-4 \times 1 \times (-4) = +16$
Square roots (Grade 8)	Understanding a square root as "the number whose square is that number", as in $\sqrt{25} = 5$ Knowing that a square root that does not come out exactly, such as $\sqrt{2}$, can be written as an approximate decimal
Quadratic equations (Algebra 1)	Being able to rewrite an equation in the form $ax^2 + bx + c = 0$ and read off the coefficients $a, b, c$ with their signs Being able to check that, for an equation you can factor, the answers match those from the quadratic formula
Graphs of quadratic functions (Algebra 1)	Knowing that the graph of $y = ax^2 + bx + c$ is a parabola Being able to picture the real solutions as "the x-coordinates of the points where the parabola meets the x-axis"

How to calculate it in Excel

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

Table to find two real solutions with the quadratic formula (when D > 0)

Coefficient a	1
Coefficient b	3
Constant term c	-4
Discriminant D = b² − 4ac	=B2^2-4B1B3
Solution 1, x₁	=(-B2+SQRT(B4))/(2*B1)
Solution 2, x₂	=(-B2-SQRT(B4))/(2*B1)

Table to find the type of solutions from the discriminant

Coefficient a	1
Coefficient b	1
Constant term c	1
Discriminant D = b² − 4ac	=B2^2-4B1B3
Type of solutions	=IF(B4>0,"Two different real solutions",IF(B4=0,"Double root","Two complex solutions"))

Table to find the double root (when D = 0)

Coefficient a	1
Coefficient b	2
Constant term c	1
Double root x	=-B2/(2*B1)

Table to find the real and imaginary parts of the two complex solutions (when D < 0)

Coefficient a	1
Coefficient b	1
Constant term c	1
Real part (−b ÷ 2a)	=-B2/(2*B1)
Imaginary part (the ± part)	=SQRT(4B1B3-B2^2)/ABS(2*B1)

Table to find the solution when a = 0 (a linear equation)

Coefficient b	2
Constant term c	4
Solution x	=-B2/B1

After pasting, the upper rows (the coefficients a, b and c) are your inputs and the lower rows are calculated automatically.
"^2" means squared, and "SQRT( )" is the function for the square root.
In the first table, for example, you see the discriminant 25, solution 1 = 1 and solution 2 = −4.
The SQRT in the first table only works when D > 0 (if D < 0 you get a #NUM! error; in that case use the fourth table, for complex solutions).

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 to find two real solutions with the quadratic formula (when D > 0)

Coefficient a	1
Coefficient b	3
Constant term c	-4
Discriminant D = b² − 4ac	=B2^2-4B1B3
Solution 1, x₁	=(-B2+SQRT(B4))/(2*B1)
Solution 2, x₂	=(-B2-SQRT(B4))/(2*B1)

Table to find the type of solutions from the discriminant

Coefficient a	1
Coefficient b	1
Constant term c	1
Discriminant D = b² − 4ac	=B2^2-4B1B3
Type of solutions	=IF(B4>0,"Two different real solutions",IF(B4=0,"Double root","Two complex solutions"))

Table to find the double root (when D = 0)

Coefficient a	1
Coefficient b	2
Constant term c	1
Double root x	=-B2/(2*B1)

Table to find the real and imaginary parts of the two complex solutions (when D < 0)

Coefficient a	1
Coefficient b	1
Constant term c	1
Real part (−b ÷ 2a)	=-B2/(2*B1)
Imaginary part (the ± part)	=SQRT(4B1B3-B2^2)/ABS(2*B1)

Table to find the solution when a = 0 (a linear equation)

Coefficient b	2
Constant term c	4
Solution x	=-B2/B1

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the coefficients with the ones from your own equation.

How to calculate it in Python

import math

a = 1.0   # coefficient of x^2
b = 3.0   # coefficient of x
c = -4.0  # constant term

if a == 0 and b == 0:
    print("If a and b are both 0, there is no equation to solve")
elif a == 0:
    # When a = 0, it becomes the linear equation bx + c = 0
    print(f"Solution of the linear equation: x = {-c / b}")
else:
    discriminant = b**2 - 4*a*c  # discriminant D
    print(f"Discriminant D = {discriminant}")
    if discriminant > 0:
        root = math.sqrt(discriminant)
        print(f"Two different real solutions: x1 = {(-b + root) / (2*a)}, x2 = {(-b - root) / (2*a)}")
    elif discriminant == 0:
        print(f"Double root: x = {-b / (2*a)}")
    else:
        real_part = -b / (2*a)
        imaginary_part = math.sqrt(-discriminant) / (2*abs(a))
        print(f"Two complex solutions: x = {real_part} ± {imaginary_part}i")

Runs with the standard library only. "**2" means squared and math.sqrt( ) is the square root. Change a, b and c at the top to the coefficients of your own equation and run it.

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

The quadratic formula

x = (−b ± √(b² − 4ac)) ÷ (2a)

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>x</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>&#x2212;</mo><mi>b</mi>
        <mo>&#xB1;</mo>
        <msqrt>
          <mrow>
            <msup><mi>b</mi><mn>2</mn></msup>
            <mo>&#x2212;</mo>
            <mn>4</mn><mi>a</mi><mi>c</mi>
          </mrow>
        </msqrt>
      </mrow>
      <mrow><mn>2</mn><mi>a</mi></mrow>
    </mfrac>
  </mrow>
</math>

x = (-b +- sqrt(b^2 - 4ac)) / (2a)

Solve[a*x^2 + b*x + c == 0, x]

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

x = roots([a b c]);

x = (-b ± √(b^2 - 4ac))/(2a)

The discriminant (tells you the type of solutions)

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

b^2 - 4*a*c

d := b^2 - 4*a*c;

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

D = b^2 - 4ac

Double root (when the discriminant $D = 0$)

x = −b ÷ (2a)

x = -\frac{b}{2a}

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

x = -b / (2a)

-b/(2*a)

x := -b/(2*a);

x = -b/(2*a);

x = -b/(2a)

Two complex solutions (when the discriminant $D < 0$)

x = (−b ± √(4ac − b²) i) ÷ (2a)

x = \frac{-b \pm \sqrt{4ac - b^2}\,i}{2a}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>x</mi>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <mo>&#x2212;</mo><mi>b</mi>
        <mo>&#xB1;</mo>
        <msqrt>
          <mrow>
            <mn>4</mn><mi>a</mi><mi>c</mi>
            <mo>&#x2212;</mo>
            <msup><mi>b</mi><mn>2</mn></msup>
          </mrow>
        </msqrt>
        <mi>i</mi>
      </mrow>
      <mrow><mn>2</mn><mi>a</mi></mrow>
    </mfrac>
  </mrow>
</math>

x = (-b +- sqrt(4ac - b^2) i) / (2a)

{(-b + Sqrt[4*a*c - b^2]*I)/(2*a), (-b - Sqrt[4*a*c - b^2]*I)/(2*a)}

x := [(-b + sqrt(4*a*c - b^2)*I)/(2*a), (-b - sqrt(4*a*c - b^2)*I)/(2*a)];

x = [(-b + sqrt(4*a*c - b^2)*1i)/(2*a), (-b - sqrt(4*a*c - b^2)*1i)/(2*a)];

x = (-b ± √(4ac - b^2) i)/(2a)

When $a = 0$ (it becomes a linear equation)

x = −c ÷ b

x = -\frac{c}{b}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>x</mi>
    <mo>=</mo>
    <mo>&#x2212;</mo>
    <mfrac><mi>c</mi><mi>b</mi></mfrac>
  </mrow>
</math>

x = -c/b

-c/b

x := -c/b;

x = -c/b;

x = -c/b

How to have ChatGPT do the calculation

You are a calculation assistant for math. Do the following calculation 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 equation x² + 3x − 4 = 0 with the quadratic formula.
Show each of the following:
1. The value of the discriminant D = b² − 4ac and the type of solutions it shows (two different real solutions, a double root, or two complex solutions)
2. The values of the solutions x (if they are complex, in the form "real part ± imaginary part i")
3. The steps showing how the values are put into the quadratic formula

Show the formulas you used and the numbers from the execution result.

How to Use

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Enter your numbers

Type the numbers you want to calculate with into the input fields
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Press the "Calculate" button
3

Check the result

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\(a, b, c\)	a, b, c	The coefficients of the quadratic equation. \(a\) is the number in front of \(x^2\), \(b\) is the number in front of \(x\), and \(c\) is the constant term (the number-only term).
\(x\)	ex	The unknown number you want to find as the solution of the equation. A quadratic equation has at most two solutions.
\(D\)	dee (discriminant)	The discriminant, \(D = b^2 - 4ac\), from the first letter of "discriminant". Some books write it with the Greek letter \(\Delta\) (delta).
\(\sqrt{\phantom{9}}\)	square root	The symbol for "the number, 0 or greater, whose square is the number inside". \(\sqrt{25} = 5\) and \(\sqrt{2} \approx 1.414\).
\(\pm\)	plus or minus	A symbol that writes "both the + case and the − case" at once. In the quadratic formula, it means that both the sum and the difference are solutions.
\(i\)	i (imaginary unit)	The number whose square is \(-1\) (\(i^2 = -1\)). It is used to write the solutions when the number under the √ is negative.
\(x_1, x_2\)	x sub 1, x sub 2	Names for the two solutions. On this page, the + side of ± is \(x_1\) and the − side is \(x_2\).

quadratic equation	An equation with an \(x^2\) (squared) term. It can be written in the form \(ax^2 + bx + c = 0\) (\(a \neq 0\)), called standard form. Taught in Algebra 1.
coefficient	The number in front of a letter. In \(3x\), the coefficient is 3. When reading coefficients, watch for hidden 1s and signs: \(x^2 - x = 0\) means a = 1, b = −1, c = 0.
solution	A value of the letter that makes the equation true. To "solve" an equation means to find all of its solutions. The solutions of a quadratic are also called its roots.
quadratic formula	A formula that calculates the solutions of a quadratic equation directly from the coefficients a, b and c. Its strength is that it always works, even when you cannot see how to factor.
discriminant	The part under the √ in the quadratic formula, \(b^2 - 4ac\). Its sign alone tells you the type of solutions (two real solutions, a double root, or two complex solutions).
double root	A solution where the two solutions come together as one. It happens when the discriminant is 0, and on the graph the parabola touches the x-axis. It is also called a repeated root.
real solution	A solution that is a real number (a number on the ordinary number line). There are real solutions when the discriminant is 0 or greater.
complex solution	A solution that is not a real number and is written with the imaginary unit \(i\). When the discriminant is negative, the solutions are a pair of the form "real part ± imaginary part \(i\)", such as \(-0.5 \pm 0.866i\) (complex conjugates). Also called imaginary solutions. Taught in Algebra 2.
completing the square	Rewriting an equation in the form \((x + p)^2 = q\) (a square equals a number). Doing this all the way with the letters a, b and c gives the quadratic formula.
factoring	Writing an expression as a product. If you can factor \(x^2 + 3x - 4 = (x - 1)(x + 4)\), you get the solutions x = 1, −4 without the quadratic formula.
parabola	The shape of the graph of \(y = ax^2 + bx + c\). The real solutions of the quadratic equation are the x-coordinates of the points where this parabola meets the x-axis (the x-intercepts).

Quadratic Formula Calculator (Discriminant and Complex Roots)

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