-2y-2=(2y-3)(y+1)

Solution for -2y-2=(2y-3)(y+1) equation:

-2y-2=(2y-3)(y+1)

We move all terms to the left:

-2y-2-((2y-3)(y+1))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2y^2+2y-3y-3)), so:

(+2y^2+2y-3y-3)

We get rid of parentheses

2y^2+2y-3y-3

We add all the numbers together, and all the variables

2y^2-1y-3

Back to the equation:

-(2y^2-1y-3)


We add all the numbers together, and all the variables

-2y-(2y^2-1y-3)-2=0

We get rid of parentheses

-2y^2-2y+1y+3-2=0

We add all the numbers together, and all the variables

-2y^2-1y+1=0

a = -2; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·(-2)·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*-2}=\frac{-2}{-4}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*-2}=\frac{4}{-4}
=-1
$