(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 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^2-10y+17=0

a = 1; b = -10; c = +17;
Δ = b2-4ac
Δ = -102-4·1·17
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{2}}{2*1}=\frac{10-4\sqrt{2}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{2}}{2*1}=\frac{10+4\sqrt{2}}{2}
$