Answer
$y=-4\pm\sqrt{2}i$
Work Step by Step
$y^{2}+8y+18=0$
Take the $18$ to the right side of the equation:
$y^{2}+8y=-18$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=8$
$y^{2}+8y+\Big(\dfrac{8}{2}\Big)^{2}=-18+\Big(\dfrac{8}{2}\Big)^{2}$
$y^{2}+8y+16=-2$
Factor the expression on the left side of the equation, which is a perfect square trinomial:
$(y+4)^{2}=-2$
Take the square root of both sides of the equation:
$\sqrt{(y+4)^{2}}=\sqrt{-2}$
$y+4=\pm\sqrt{2}i$
Solve for $y$"
$y=-4\pm\sqrt{2}i$