2y+y^2-128=0

Solution for 2y+y^2-128=0 equation:

2y+y^2-128=0

a = 1; b = 2; c = -128;
Δ = b2-4ac
Δ = 22-4·1·(-128)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{129}}{2*1}=\frac{-2-2\sqrt{129}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{129}}{2*1}=\frac{-2+2\sqrt{129}}{2}
$