# How do you solve x^2+x-6=0 by completing the square?

Jul 2, 2017

$x = 2$ or $- 3$

#### Explanation:

Moving the $- 6$ to the right side:
${x}^{2} + x = 6$

Completing the square by adding ${\left(\frac{b}{2}\right)}^{2}$ to both sides, where $b$ is the second coefficient (in this case, $b = 1$):
${x}^{2} + x + {\left(\frac{1}{2}\right)}^{2} = 6 + {\left(\frac{1}{2}\right)}^{2}$
${x}^{2} + x + \frac{1}{4} = 6 + \frac{1}{4}$
${\left(x + \frac{1}{2}\right)}^{2} = \frac{25}{4}$

Simplifying:
$x + \frac{1}{2} = \setminus \pm \frac{5}{2}$
$x = - \frac{1}{2} \setminus \pm \frac{5}{2}$
$x = \frac{4}{2}$ or $- \frac{6}{2}$
$x = 2$ or $- 3$