#### Answer

$y=1$ and $y=-2$

#### Work Step by Step

$y^{2}+y-2=0$
Take the $2$ to the right side of the equation:
$y^{2}+y=2$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=1$
$y^{2}+y+\Big(\dfrac{1}{2}\Big)^{2}=2+\Big(\dfrac{1}{2}\Big)^{2}$
$y^{2}+y+\dfrac{1}{4}=\dfrac{9}{4}$
Factor the expression on the left side of the equation, which is a perfect square trinomial:
$\Big(y+\dfrac{1}{2}\Big)^{2}=\dfrac{9}{4}$
Take the square root of both sides of the equation:
$\sqrt{\Big(y+\dfrac{1}{2}\Big)^{2}}=\sqrt{\dfrac{9}{4}}$
$y+\dfrac{1}{2}=\pm\dfrac{3}{2}$
Solve for $y$:
$y=\dfrac{-1\pm3}{2}$
The two solutions are:
$y=\dfrac{-1+3}{2}=1$
$y=\dfrac{-1-3}{2}=-2$