# How do you solve x^2=9 using the quadratic formula?

Feb 27, 2016

$x = \pm 3$

#### Explanation:

Write as :$\text{ } {x}^{2} - 9 = 0$

Compare to standard form of:$\text{ } a {x}^{2} + b x + c = 0$

Where:$\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 1$
$b = 0$
$c = - 9$

So by substitution you have:

$x = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \left(1\right) \left(- 9\right)}}{2 \times 1}$

$x = \pm \frac{\sqrt{+ 36}}{2}$

$x = \pm \frac{\sqrt{{6}^{2}}}{2}$

$x = \pm 3$