4y(2y-4)=16y+4y^2

Solution for 4y(2y-4)=16y+4y^2 equation:

4y(2y-4)=16y+4y^2

We move all terms to the left:

4y(2y-4)-(16y+4y^2)=0

We multiply parentheses

-(16y+4y^2)+8y^2-16y=0

We get rid of parentheses

-4y^2+8y^2-16y-16y=0

We add all the numbers together, and all the variables

4y^2-32y=0

a = 4; b = -32; c = 0;
Δ = b2-4ac
Δ = -322-4·4·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32}{2*4}=\frac{0}{8}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32}{2*4}=\frac{64}{8}
=8
$