(2y-8)(y+5)*2=(2y+8)(2y-8)

Solution for (2y-8)(y+5)*2=(2y+8)(2y-8) equation:

(2y-8)(y+5)*2=(2y+8)(2y-8)

We move all terms to the left:

(2y-8)(y+5)*2-((2y+8)(2y-8))=0

We use the square of the difference formula

4y^2+(2y-8)(y+5)*2+64=0

We multiply parentheses ..

4y^2+(+2y^2+10y-8y-40)*2+64=0

We multiply parentheses

4y^2+4y^2+20y-16y-80+64=0

We add all the numbers together, and all the variables

8y^2+4y-16=0

a = 8; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·8·(-16)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{33}}{2*8}=\frac{-4-4\sqrt{33}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{33}}{2*8}=\frac{-4+4\sqrt{33}}{16}
$