(y+3)(y-5)=2y-2,

Solution for (y+3)(y-5)=2y-2, equation:

(y+3)(y-5)=2y-2.

We move all terms to the left:

(y+3)(y-5)-(2y-2.)=0

We add all the numbers together, and all the variables

(y+3)(y-5)-(2y-2)=0

We get rid of parentheses

(y+3)(y-5)-2y+2=0

We multiply parentheses ..

(+y^2-5y+3y-15)-2y+2=0

We get rid of parentheses

y^2-5y+3y-2y-15+2=0

We add all the numbers together, and all the variables

y^2-4y-13=0

a = 1; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·1·(-13)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{17}}{2*1}=\frac{4-2\sqrt{17}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{17}}{2*1}=\frac{4+2\sqrt{17}}{2}
$