2y^2-4y-8=-y^2=y

Solution for 2y^2-4y-8=-y^2=y equation:

2y^2-4y-8=-y^2=y

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3y^2-4y-8=0

a = 3; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·3·(-8)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*3}=\frac{4-4\sqrt{7}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*3}=\frac{4+4\sqrt{7}}{6}
$