# How do you solve by completing the square 2x^2+4x+12=0?

Apr 1, 2015

Start by writing:
$2 {x}^{2} + 4 x = - 12$
${x}^{2} + \frac{4}{2} x = - \frac{12}{2}$
${x}^{2} + 2 x = - 6$
${x}^{2} + 2 x + 1 - 1 = - 6$
${\left(x + 1\right)}^{2} = - 6 + 1$
${\left(x + 1\right)}^{2} = - 5$
$\left(x + 1\right) = \pm \sqrt{- 5}$
You have a negative squre root!
Did you see complex numbers yet?
If NOT stop here and say "no real solutions".
If YES:
$x = - 1 \pm i \sqrt{5}$