x^2+20x=-20

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

x^2+20x=-20

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+20x+20=0

a = 1; b = 20; c = +20;
Δ = b2-4ac
Δ = 202-4·1·20
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*1}=\frac{-20-8\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*1}=\frac{-20+8\sqrt{5}}{2}
$