Quadratic Formula

The Formula

For any quadratic $ax^2 + bx + c = 0$:

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

Plug in $a$, $b$, $c$. Get the answers.

Where Does It Come From?

It’s completing the square on the general form, done once for all quadratics.

Start with the general quadratic:

$ax^2 + bx + c = 0$

Divide everything by $a$:

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

Move the constant to the right:

$x^2 + \frac{b}{a}x = -\frac{c}{a}$

Add $\left(\dfrac{b}{2a}\right)^2$ to both sides to complete the square:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$

Left side is now a perfect square:

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

Take the square root:

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

Solve for $x$:

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

Example 1

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

Here $a = 1$, $b = 5$, $c = 6$.

So $x = \dfrac{-5 + 1}{2} = -2$ or $x = \dfrac{-5 - 1}{2} = -3$.

Example 2

Solve: $2x^2 - 4x - 6 = 0$

Here $a = 2$, $b = -4$, $c = -6$.

So $x = \dfrac{12}{4} = 3$ or $x = \dfrac{-4}{4} = -1$.

Example 3: Irrational Roots

Solve: $x^2 - 2x - 1 = 0$

Here $a = 1$, $b = -2$, $c = -1$.

When to Use It

The quadratic formula always works.