Answer
$a=-2\pm\sqrt{2}i$
Work Step by Step
$2a^{2}+8a=-12$
Take out common factor $2$ from the left side of the equation:
$2(a^{2}+4a)=-12$
Take the $2$ to divide the right side of the equation:
$a^{2}+4a=\dfrac{-12}{2}$
$a^{2}+4a=-6$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular case, $b=4$
$a^{2}+4a+\Big(\dfrac{4}{2}\Big)^{2}=-6+\Big(\dfrac{4}{2}\Big)^{2}$
$a^{2}+4a+4=-6+4$
$a^{2}+4a+4=-2$
Factor the left side of the equation, which is a perfect square trinomial:
$(a+2)^{2}=-2$
Take the square root of both sides of the equation:
$\sqrt{(a+2)^{2}}=\sqrt{-2}$
$a+2=\pm\sqrt{2}i$
Solve for $a$:
$a=-2\pm\sqrt{2}i$
