-3y=2y^2-2

Solution for -3y=2y^2-2 equation:

-3y=2y^2-2

We move all terms to the left:

-3y-(2y^2-2)=0

We get rid of parentheses

-2y^2-3y+2=0

a = -2; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·(-2)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-2}=\frac{-2}{-4}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-2}=\frac{8}{-4}
=-2
$