# How do you solve the equation x^2-3x-18=0 by using the quadratic formula?

May 11, 2018

$x = - 3 \mathmr{and} x = 6$

#### Explanation:

It is really worth trying to commit this formula to memory.

Given: ${x}^{2} - 3 x - 18 = 0$

Compare to $a {x}^{2} + b x + c = 0$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Comparing the two we have:

a=1; b=-3 and c=-18

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 18\right)}}{2 \left(1\right)}$

$x = \frac{3 \pm \sqrt{81}}{2 \left(1\right)}$

$x = \frac{3 \pm 9}{2}$

$x = - 3 \mathmr{and} x = 6$