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

${x}^{2} + x = 6$ add and subtract $\frac{1}{4}$ on the left side:
${x}^{2} + x + \frac{1}{4} - \frac{1}{4} = 6$ take $- \frac{1}{4}$ to the right:
${x}^{2} + x + \frac{1}{4} = 6 + \frac{1}{4}$
${\left(x + \frac{1}{2}\right)}^{2} = 6 + \frac{1}{4}$
${\left(x + \frac{1}{2}\right)}^{2} = \frac{25}{4}$ square root both sides:
$x + \frac{1}{2} = \pm \frac{5}{2}$
${x}_{1} = - \frac{1}{2} + \frac{5}{2} = 2$
${x}_{2} = - \frac{1}{2} - \frac{5}{2} = - 3$