-y^2+10y+8=0

Solution for -y^2+10y+8=0 equation:

-y^2+10y+8=0

We add all the numbers together, and all the variables

-1y^2+10y+8=0

a = -1; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-1)·8
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*-1}=\frac{-10-2\sqrt{33}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*-1}=\frac{-10+2\sqrt{33}}{-2}
$