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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

y^2-y-2y-8=0

We add all the numbers together, and all the variables

y^2-3y-8=0

a = 1; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·1·(-8)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*1}=\frac{3-\sqrt{41}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*1}=\frac{3+\sqrt{41}}{2}
$