2y^2=-3y+2

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

2y^2=-3y+2

We move all terms to the left:

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