Quadratic Formula: Steps and Examples

The quadratic formula solves any quadratic equation $ax^2 + bx + c = 0$ . The solutions are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Put the equation in standard form, find $a$ , $b$ and $c$ , and plug them in. It works even when the quadratic won’t factor. The answers are where the graph crosses the x-axis.

Before this lesson: You should know completing the square, solving quadratic equations by factoring and simplifying square roots.

What Does the Quadratic Formula Do?

Some quadratics factor nicely. Many don’t. The quadratic formula gives the answers to every quadratic, so you never get stuck.

The answers are the x-intercepts of the graph of $y = ax^2 + bx + c$ . The graph below shows $y = x^2 - 2x - 4$ . It crosses the x-axis at $x = 1 \pm \sqrt{5}$ , which don’t come from factoring.

The parabola y = x² − 2x − 4 on a grid from −3 to 5 across and −6 to 2 up. Its vertex is (1, −5). It crosses the x-axis at about (−1.24, 0) and (3.24, 0), the solutions x = 1 ± √5 from the quadratic formula. Both points are 2.24 units from the dashed axis of symmetry x = 1. — The quadratic formula gives x = 1 ± √5, about 3.24 and −1.24. These are where the parabola crosses the x-axis.

The Quadratic Formula

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

Here $a$ , $b$ and $c$ come from $ax^2 + bx + c = 0$ , and $a \ne 0$ . The $\pm$ sign means you get two answers: one with + and one with −.

The part under the square root, $b^2 - 4ac$ , is called the discriminant. It tells you how many real solutions there are.

Steps for using the quadratic formula

Write the equation in standard form: $ax^2 + bx + c = 0$ .
Find $a$ , $b$ and $c$ . Keep their signs.
Find $b^2 - 4ac$ first.
Plug everything into the formula.
Simplify the square root. Then find both answers.

Worked Examples

Example 1: Two whole-number answers

Solve $x^2 + 5x + 6 = 0$ .

Here $a = 1$ , $b = 5$ and $c = 6$ . First, $b^2 - 4ac = 25 - 24 = 1$ .

x = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}

So $x = \frac{-4}{2} = -2$ or $x = \frac{-6}{2} = -3$ . You could also factor this one: $(x + 2)(x + 3) = 0$ .

Example 2: Answers with a square root

Solve $x^2 - 2x - 4 = 0$ .

Here $a = 1$ , $b = -2$ and $c = -4$ . So $-b = 2$ , and $b^2 - 4ac = 4 + 16 = 20$ .

\begin{aligned} x &= \frac{2 \pm \sqrt{20}}{2} \\ &= \frac{2 \pm 2\sqrt{5}}{2} \\ &= 1 \pm \sqrt{5} \end{aligned}

Using $\sqrt{5} \approx 2.24$ , the answers are about 3.24 and −1.24. They match the graph above.

Example 3: When a is not 1

Solve $2x^2 + 3x - 5 = 0$ .

Here $a = 2$ , $b = 3$ and $c = -5$ . So $b^2 - 4ac = 9 + 40 = 49$ , and $2a = 4$ .

x = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}

So $x = \frac{4}{4} = 1$ or $x = \frac{-10}{4} = -\frac{5}{2}$ .

Example 4: Write it in standard form first

Solve $3x^2 = 4x + 2$ .

Subtract $4x$ and 2 from both sides: $3x^2 - 4x - 2 = 0$ . Now $a = 3$ , $b = -4$ and $c = -2$ . So $b^2 - 4ac = 16 + 24 = 40$ .

\begin{aligned} x &= \frac{4 \pm \sqrt{40}}{6} \\ &= \frac{4 \pm 2\sqrt{10}}{6} \\ &= \frac{2 \pm \sqrt{10}}{3} \end{aligned}

To simplify, divide every term on top and the bottom by 2.

Example 5: Word problem

A ball is thrown up from 6 feet high. Its height after $t$ seconds is $h = -16t^2 + 32t + 6$ feet. When does it hit the ground? Use $\sqrt{22} \approx 4.69$ .

The ball hits the ground when $h = 0$ , so solve $-16t^2 + 32t + 6 = 0$ . Here $b^2 - 4ac = 1024 + 384 = 1408$ . Since $1408 = 64 \cdot 22$ , $\sqrt{1408} = 8\sqrt{22}$ .

\begin{aligned} t &= \frac{-32 \pm 8\sqrt{22}}{-32} \\ &= 1 \mp \frac{\sqrt{22}}{4} \end{aligned}

Time can’t be negative, so use the plus answer: $t = 1 + \frac{\sqrt{22}}{4} \approx 1 + \frac{4.69}{4} \approx 2.17$ . The ball hits the ground after about 2.17 seconds.

Common Mistakes

Mistake	Example of the error	Fix
Wrong sign on −b	For $b = -2$ , writing $-b = -2$	$-b$ is the opposite of $b$ : $-(-2) = 2$ .
Squaring a negative wrong	$(-4)^2 = -16$	A negative squared is positive: $(-4)^2 = 16$ .
Dividing only part by 2a	$-b \pm \frac{\sqrt{b^2 - 4ac}}{2a}$	The fraction bar goes under the whole top, $-b \pm \sqrt{b^2 - 4ac}$ .
Not in standard form first	For $3x^2 = 4x + 2$ , using $c = 2$	Move every term to one side first: $3x^2 - 4x - 2 = 0$ , so $c = -2$ .
Using 2 instead of 2a	For $2x^2 + 3x - 5 = 0$ , dividing by 2	The bottom is $2a = 4$ .

Which Method Should I Use?

Every quadratic can be solved with the quadratic formula. But a faster way often works.

Equation looks like	Best method	Example
No x term	Square roots	$x^2 = 49$
Factors easily	Factoring	$x^2 + 5x + 6 = 0$
$a = 1$ and $b$ is even	Completing the square	$x^2 + 6x - 7 = 0$
Anything else	Quadratic formula	$3x^2 - 4x - 2 = 0$

The formula comes from completing the square on $ax^2 + bx + c = 0$ . It does the same steps once, for every quadratic.

Practice Problems

Solve $x^2 + 7x + 10 = 0$ .
Solve $x^2 - 3x - 10 = 0$ .
Solve $2x^2 - 7x + 3 = 0$ .
Solve $x^2 + 4x - 1 = 0$ .
Solve $x^2 - 6x + 4 = 0$ .
Solve $2x^2 + 2x - 1 = 0$ .
Solve $5x^2 - x = 4$ .
Solve $4x^2 = 12x - 9$ .
Solve $x^2 + 2x + 5 = 0$ .
A garden is 2 meters longer than it is wide. Its area is 50 square meters. Find its width to the nearest hundredth. Use $\sqrt{51} \approx 7.14$ .

Show answers

$b^2 - 4ac = 49 - 40 = 9$ , so $x = \frac{-7 \pm 3}{2}$ . $x = -2$ or $x = -5$ .
$b^2 - 4ac = 9 + 40 = 49$ , so $x = \frac{3 \pm 7}{2}$ . $x = 5$ or $x = -2$ .
$b^2 - 4ac = 49 - 24 = 25$ , so $x = \frac{7 \pm 5}{4}$ . $x = 3$ or $x = \frac{1}{2}$ .
$b^2 - 4ac = 16 + 4 = 20$ , so $x = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$ .
$b^2 - 4ac = 36 - 16 = 20$ , so $x = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$ .
$b^2 - 4ac = 4 + 8 = 12$ , so $x = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$ .
Standard form: $5x^2 - x - 4 = 0$ . $b^2 - 4ac = 1 + 80 = 81$ , so $x = \frac{1 \pm 9}{10}$ . $x = 1$ or $x = -\frac{4}{5}$ .
Standard form: $4x^2 - 12x + 9 = 0$ . $b^2 - 4ac = 144 - 144 = 0$ , so $x = \frac{12}{8} = \frac{3}{2}$ . There is only one answer.
$b^2 - 4ac = 4 - 20 = -16$ . You can’t take the square root of a negative number, so there are no real solutions.
$w(w + 2) = 50$ , so $w^2 + 2w - 50 = 0$ . $b^2 - 4ac = 4 + 200 = 204$ , and $\sqrt{204} = 2\sqrt{51}$ . So $w = -1 \pm \sqrt{51}$ . A width must be positive: $w \approx -1 + 7.14 = 6.14$ meters.

FAQ

What is the quadratic formula?

It is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . It gives the solutions of any quadratic equation $ax^2 + bx + c = 0$ .

When should you use the quadratic formula?

Use it when a quadratic doesn’t factor easily, or when you aren’t sure. It always works.

What if the number under the square root is negative?

Then the equation has no real solutions. The graph never crosses the x-axis.

How do you remember the quadratic formula?

Many students sing it to the tune of “Pop Goes the Weasel.” Writing it out a few times while you practice also helps.

Summary

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Write the equation in standard form before you find $a$ , $b$ and $c$ .
Find $b^2 - 4ac$ first, and watch the signs.
Simplify the square root, then find both answers.
It works for every quadratic, even ones that don’t factor.

Previous lesson: Completing the Square

Next lesson: The Discriminant

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