(4y-2)(2y-2)=12

Solution for (4y-2)(2y-2)=12 equation:

(4y-2)(2y-2)=12

We move all terms to the left:

(4y-2)(2y-2)-(12)=0

We multiply parentheses ..

(+8y^2-8y-4y+4)-12=0

We get rid of parentheses

8y^2-8y-4y+4-12=0

We add all the numbers together, and all the variables

8y^2-12y-8=0

a = 8; b = -12; c = -8;
Δ = b2-4ac
Δ = -122-4·8·(-8)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*8}=\frac{-8}{16}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*8}=\frac{32}{16}
=2
$