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

Solution for (2y+2)(3y-3)=0 equation:

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

We multiply parentheses ..

(+6y^2-6y+6y-6)=0

We get rid of parentheses

6y^2-6y+6y-6=0

We add all the numbers together, and all the variables

6y^2-6=0

a = 6; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·6·(-6)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*6}=\frac{-12}{12}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*6}=\frac{12}{12}
=1
$