# How do you solve 2x^2 + 32x + 12 = 0 using completing the square?

Jun 24, 2015

I found:
${x}_{1} = - 8 + \sqrt{58}$
${x}_{2} = - 8 - \sqrt{58}$

#### Explanation:

Write it as (dividing by $2$):
${x}^{2} / 2 + \frac{32}{2} x + \frac{12}{2} = 0$
${x}^{2} + 16 x + 6 = 0$
${x}^{2} + 16 x = - 6$
Add and subtract $64$:
${x}^{2} + 16 x + 64 - 64 = - 6$
${x}^{2} + 16 x + 64 = - 6 + 64$
${\left(x + 8\right)}^{2} = 58$
$x + 8 = \pm \sqrt{58}$
$x = - 8 \pm \sqrt{58}$
so:
${x}_{1} = - 8 + \sqrt{58}$
${x}_{2} = - 8 - \sqrt{58}$