x^2+20x-20=-116

Solution for x^2+20x-20=-116 equation:

x^2+20x-20=-116

We move all terms to the left:

x^2+20x-20-(-116)=0

We add all the numbers together, and all the variables

x^2+20x+96=0

a = 1; b = 20; c = +96;
Δ = b2-4ac
Δ = 202-4·1·96
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*1}=\frac{-16}{2}
=-8
$