y^2=2y+99

Solution for y^2=2y+99 equation:

y^2=2y+99

We move all terms to the left:

y^2-(2y+99)=0

We get rid of parentheses

y^2-2y-99=0

a = 1; b = -2; c = -99;
Δ = b2-4ac
Δ = -22-4·1·(-99)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-20}{2*1}=\frac{-18}{2}
=-9
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+20}{2*1}=\frac{22}{2}
=11
$