-4y+8=4(2y+2);y

Solution for -4y+8=4(2y+2);y equation:

-4y+8=4(2y+2)y

We move all terms to the left:

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


We calculate terms in parentheses: -(4(2y+2)y), so:

4(2y+2)y

We multiply parentheses

8y^2+8y

Back to the equation:

-(8y^2+8y)


We get rid of parentheses

-8y^2-4y-8y+8=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
$