Answer
$y=-1\pm i$
Work Step by Step
$y^{2}+2y+2=0$
Take the $2$ to the right side of the equation:
$y^{2}+2y=-2$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular case, $b=2$
$y^{2}+2y+\Big(\dfrac{2}{2}\Big)^{2}=-2+\Big(\dfrac{2}{2}\Big)^{2}$
$y^{2}+2y+1=-2+1$
$y^{2}+2y+1=-1$
Factor the left side of the equation, which is a perfect square trinomial:
$(y+1)^{2}=-1$
Take the square root of both sides of the equation:
$\sqrt{(y+1)^{2}}=\sqrt{-1}$
$y+1=\pm i$
Solve for $y$:
$y=-1\pm i$
