The Quadratic Formula

Factoring is quick, but many quadratic equations don’t factor nicely: try finding two integers that multiply to $-4$ and add to $2$. The quadratic formula solves every quadratic equation, whether it factors or not. It comes from completing the square once, in general, so you never have to do it again.

Key ideas

The formula

For any quadratic equation $ax^2 + bx + c = 0$ with $a \ne 0$:

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

The $\pm$ means there are usually two answers: one using $+$ and one using $-$.

Where it comes from

The formula comes from completing the square on $ax^2 + bx + c = 0$. You don’t need to reproduce this general development yourself, but it’s worth following once so the formula isn’t magic. On the left is a numerical example, $2x^2 + 8x + 3 = 0$; on the right, the same steps with letters.

Step	Example	General case
Start	$2x^2 + 8x + 3 = 0$	$ax^2 + bx + c = 0$
Divide by $a$	$x^2 + 4x + \tfrac{3}{2} = 0$	$x^2 + \tfrac{b}{a}x + \tfrac{c}{a} = 0$
Move the constant	$x^2 + 4x = -\tfrac{3}{2}$	$x^2 + \tfrac{b}{a}x = -\tfrac{c}{a}$
Add (half of the $x$-coefficient)² to both sides	$x^2 + 4x + 4 = 4 - \tfrac{3}{2}$	$x^2 + \tfrac{b}{a}x + \tfrac{b^2}{4a^2} = \tfrac{b^2}{4a^2} - \tfrac{c}{a}$
Write as a square	$(x + 2)^2 = \tfrac{5}{2}$	$\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2}$
Square root both sides	$x + 2 = \pm\sqrt{\tfrac{5}{2}}$	$x + \tfrac{b}{2a} = \pm\tfrac{\sqrt{b^2 - 4ac}}{2a}$
Solve for $x$	$x = -2 \pm \sqrt{\tfrac{5}{2}}$	$x = \tfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In the “write as a square” step of the general case, the right side was put over the common denominator $4a^2$: $\tfrac{b^2}{4a^2} - \tfrac{4ac}{4a^2} = \tfrac{b^2 - 4ac}{4a^2}$.

Now check that the formula gives the same answer for the example, with $a = 2$, $b = 8$, $c = 3$:

$$x = \frac{-8 \pm \sqrt{8^2 - 4(2)(3)}}{2(2)} = \frac{-8 \pm \sqrt{40}}{4}$$

Both give $x \approx -0.42$ or $x \approx -3.58$. ✓

Using the formula

1. Rearrange the equation into the form $ax^2 + bx + c = 0$.
2. Write down $a$, $b$ and $c$, including their signs.
3. Substitute, using brackets around negative numbers.
4. Simplify $b^2 - 4ac$ (the part under the square root) first.
5. Write the two exact answers, then round to decimals if the question asks.

An exact answer keeps the square root, like $-1 + \sqrt{5}$. A decimal answer is rounded, like $1.24$. If $b^2 - 4ac$ is a perfect square, the roots are rational and the equation could also have been solved by factoring.

Choosing a method

Method	Best when	Notes
Factoring	the quadratic factors easily	fastest; gives exact roots
Quadratic formula	it doesn’t factor, or you can’t see how	always works; gives exact roots
Graphing	you want to see the roots or check them	with technology (such as Desmos), read the $x$-intercepts; usually approximate

A good habit: try factoring for a few seconds; if it doesn’t work, use the formula; then check with a graph if you can.

When b² − 4ac is negative

The expression $b^2 - 4ac$ is called the discriminant. If it’s negative, the formula asks for the square root of a negative number, and no real number squares to give a negative. So the equation has no real roots, and the graph of $y = ax^2 + bx + c$ doesn’t cross the $x$-axis at all. You’ll use the discriminant to count roots in zeros and the discriminant.

On the SAT

On the SAT, graph $y = ax^2 + bx + c$ in Desmos and click the x-intercepts to get the roots as decimals. Answer choices are often exact, so convert them: for $2x^2 + 3x - 7 = 0$, Desmos shows $x \approx 1.266$ and $x \approx -2.766$, which match $\dfrac{-3 \pm \sqrt{65}}{4}$. If the parabola doesn’t cross the x-axis, there are no real roots ($b^2 - 4ac \lt 0$). When a question asks for a root in exact form, use the formula by hand. See using Desmos on the SAT.

Worked examples

Example 1: An equation that also factors

Solve $x^2 - 3x - 10 = 0$ using the quadratic formula. Check by factoring.

Solution. $a = 1$, $b = -3$, $c = -10$.

$$\begin{aligned} x &= \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} \\ &= \frac{3 \pm \sqrt{9 + 40}}{2} \\ &= \frac{3 \pm \sqrt{49}}{2} \\ &= \frac{3 \pm 7}{2} \end{aligned}$$

So $x = \dfrac{3 + 7}{2} = 5$ or $x = \dfrac{3 - 7}{2} = -2$.

Check by factoring: $x^2 - 3x - 10 = (x - 5)(x + 2)$, which gives the same roots. ✓ Since $49$ is a perfect square, factoring was possible.

Example 2: Exact and decimal answers

Solve $x^2 + 2x - 4 = 0$. Give exact answers and decimals to two places.

Solution. $a = 1$, $b = 2$, $c = -4$.

$$\begin{aligned} x &= \frac{-2 \pm \sqrt{2^2 - 4(1)(-4)}}{2(1)} \\ &= \frac{-2 \pm \sqrt{4 + 16}}{2} \\ &= \frac{-2 \pm \sqrt{20}}{2} \end{aligned}$$

The exact roots are $x = \dfrac{-2 + \sqrt{20}}{2}$ and $x = \dfrac{-2 - \sqrt{20}}{2}$. (If you know how to simplify radicals, $\sqrt{20} = 2\sqrt{5}$, and these become $-1 \pm \sqrt{5}$.)

With a calculator, $\sqrt{20} \approx 4.472$:

$$x \approx \frac{-2 + 4.472}{2} \approx 1.24 \qquad \text{or} \qquad x \approx \frac{-2 - 4.472}{2} \approx -3.24$$

The graph of $y = x^2 + 2x - 4$ crosses the $x$-axis at the two roots, confirming the answer.

Example 3: Rearranging first

Solve $3x^2 = 5 - 4x$. Round to two decimal places.

Solution. Get $0$ on one side by adding $4x$ and subtracting $5$:

$$3x^2 + 4x - 5 = 0$$

So $a = 3$, $b = 4$, $c = -5$.

$$\begin{aligned} x &= \frac{-4 \pm \sqrt{4^2 - 4(3)(-5)}}{2(3)} \\ &= \frac{-4 \pm \sqrt{16 + 60}}{6} \\ &= \frac{-4 \pm \sqrt{76}}{6} \end{aligned}$$

With $\sqrt{76} \approx 8.718$:

$$x \approx \frac{-4 + 8.718}{6} \approx 0.79 \qquad \text{or} \qquad x \approx \frac{-4 - 8.718}{6} \approx -2.12$$

Check $x \approx 0.79$ in the original: $3(0.79)^2 \approx 1.87$ and $5 - 4(0.79) = 1.84$. These are close; the small difference comes from rounding. ✓

Example 4: No real roots

Solve $x^2 - 4x + 7 = 0$.

Solution. $a = 1$, $b = -4$, $c = 7$. Start with the part under the square root:

$$b^2 - 4ac = (-4)^2 - 4(1)(7) = 16 - 28 = -12$$

The formula would need $\sqrt{-12}$, which isn’t a real number. So the equation has no real roots.

This makes sense graphically. Completing the square gives $y = x^2 - 4x + 7 = (x - 2)^2 + 3$: the parabola opens up from its vertex $(2, 3)$, which is above the $x$-axis, so it never crosses the axis.

Common mistakes

Using the formula before the equation equals $0$. For $3x^2 = 5 - 4x$, rearrange to $3x^2 + 4x - 5 = 0$ first. Otherwise you’ll use the wrong values of $b$ and $c$.

Dropping the sign of $b$ or $c$. In $x^2 - 3x - 10 = 0$, $b = -3$ and $c = -10$. Then $-b = 3$, not $-3$.

Squaring a negative $b$ incorrectly. If $b = -4$, then $b^2 = (-4)^2 = 16$. On a calculator, type the brackets: $(-4)^2$, not $-4^2$.

Dividing only part of the top by $2a$. The whole numerator, $-b \pm \sqrt{b^2 - 4ac}$, is divided by $2a$. Draw the fraction bar all the way across.

Rounding too early. Keep the full calculator value of the square root until the last step, then round.

Writing “no solution” too quickly. A negative $b^2 - 4ac$ means no real roots. Double-check the signs in $b^2 - 4ac$ before you conclude that.

Practice

1. (Warm-up) State the values of $a$, $b$ and $c$ for each equation. Rearrange first if needed.

• (a) $4x^2 - x + 7 = 0$
• (b) $2x^2 = 9 - 3x$
• (c) $x^2 - 5 = 0$

Solution

(a) $a = 4$, $b = -1$, $c = 7$.

(b) Rearrange: $2x^2 + 3x - 9 = 0$, so $a = 2$, $b = 3$, $c = -9$.

(c) There’s no $x$ term, so $a = 1$, $b = 0$, $c = -5$.

2. (Warm-up) Solve $x^2 + 6x + 5 = 0$ using the quadratic formula.

Solution

$a = 1$, $b = 6$, $c = 5$:

$$x = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2}$$

So $x = -1$ or $x = -5$.

3. (Core) Solve $x^2 - 6x + 4 = 0$. Give exact answers and decimals to two places.

Solution

$a = 1$, $b = -6$, $c = 4$:

$$x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2}$$

Exact: $x = \dfrac{6 \pm \sqrt{20}}{2}$, which simplifies to $3 \pm \sqrt{5}$.

Decimals: $x \approx 5.24$ or $x \approx 0.76$.

4. (Core) Solve $2x^2 + 3x - 7 = 0$. Round to two decimal places.

Solution

$a = 2$, $b = 3$, $c = -7$:

$$x = \frac{-3 \pm \sqrt{9 - 4(2)(-7)}}{4} = \frac{-3 \pm \sqrt{9 + 56}}{4} = \frac{-3 \pm \sqrt{65}}{4}$$

With $\sqrt{65} \approx 8.062$: $x \approx \dfrac{5.062}{4} \approx 1.27$ or $x \approx \dfrac{-11.062}{4} \approx -2.77$.

5. (Core) Solve $-x^2 + 4x + 3 = 0$. Give exact answers and decimals to two places.

Solution

Multiplying both sides by $-1$ makes $a$ positive, which is easier to work with (it doesn’t change the roots):

$$x^2 - 4x - 3 = 0$$

$a = 1$, $b = -4$, $c = -3$:

$$x = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2}$$

Exact: $x = \dfrac{4 \pm \sqrt{28}}{2}$, which simplifies to $2 \pm \sqrt{7}$.

Decimals: $x \approx 4.65$ or $x \approx -0.65$.

6. (Core) Choose a method for each equation, explain your choice, and solve. Round to two decimal places where needed.

• (a) $x^2 - 81 = 0$
• (b) $x^2 + 5x - 3 = 0$

Solution

(a) Factoring: it’s a difference of squares. $(x - 9)(x + 9) = 0$, so $x = 9$ or $x = -9$.

(b) No two integers multiply to $-3$ and add to $5$, so use the formula with $a = 1$, $b = 5$, $c = -3$:

$$x = \frac{-5 \pm \sqrt{25 + 12}}{2} = \frac{-5 \pm \sqrt{37}}{2}$$

So $x \approx 0.54$ or $x \approx -5.54$.

7. (Core) Solve each equation, or explain why it has no real roots. Round to two decimal places.

• (a) $5x^2 - 2x = 1$
• (b) $2x^2 + 5x + 4 = 0$

Solution

(a) Rearrange: $5x^2 - 2x - 1 = 0$, so $a = 5$, $b = -2$, $c = -1$.

$$x = \frac{2 \pm \sqrt{4 + 20}}{10} = \frac{2 \pm \sqrt{24}}{10}$$

So $x \approx 0.69$ or $x \approx -0.29$.

(b) $b^2 - 4ac = 25 - 4(2)(4) = 25 - 32 = -7$. This is negative, so there are no real roots. The graph of $y = 2x^2 + 5x + 4$ never crosses the $x$-axis.

8. (Challenge) Solve $0.5x^2 - 1.2x - 3 = 0$. Round to two decimal places.

Solution

$a = 0.5$, $b = -1.2$, $c = -3$:

$$b^2 - 4ac = (-1.2)^2 - 4(0.5)(-3) = 1.44 + 6 = 7.44$$$$x = \frac{1.2 \pm \sqrt{7.44}}{2(0.5)} = \frac{1.2 \pm \sqrt{7.44}}{1}$$

With $\sqrt{7.44} \approx 2.728$: $x \approx 3.93$ or $x \approx -1.53$.

(You could also multiply the equation by $10$ first to get $5x^2 - 12x - 30 = 0$, which gives the same roots.)

9. (Challenge) Solve $x^2 + 10x + 7 = 0$ by completing the square, following the steps in the table in Key ideas. Then solve it with the quadratic formula and show that the answers match.

Solution

Completing the square. Here $a = 1$, so there’s nothing to divide.

$$\begin{aligned} x^2 + 10x &= -7 \\ x^2 + 10x + 25 &= -7 + 25 \\ (x + 5)^2 &= 18 \\ x + 5 &= \pm\sqrt{18} \\ x &= -5 \pm \sqrt{18} \end{aligned}$$

Formula. $a = 1$, $b = 10$, $c = 7$:

$$x = \frac{-10 \pm \sqrt{100 - 28}}{2} = \frac{-10 \pm \sqrt{72}}{2}$$

To compare, note that $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$, so

$$x = \frac{-10 \pm 2\sqrt{18}}{2} = -5 \pm \sqrt{18}$$

The answers match: $x \approx -0.76$ or $x \approx -9.24$.