-y^2=-8y+4

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

-y^2=-8y+4

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1y^2-(-8y+4)=0

We get rid of parentheses

-1y^2+8y-4=0

a = -1; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·(-1)·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*-1}=\frac{-8-4\sqrt{3}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*-1}=\frac{-8+4\sqrt{3}}{-2}
$