0=x^2+20x+80

Solution for 0=x^2+20x+80 equation:

0=x^2+20x+80

We move all terms to the left:

0-(x^2+20x+80)=0

We add all the numbers together, and all the variables

-(x^2+20x+80)=0

We get rid of parentheses

-x^2-20x-80=0

We add all the numbers together, and all the variables

-1x^2-20x-80=0

a = -1; b = -20; c = -80;
Δ = b2-4ac
Δ = -202-4·(-1)·(-80)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*-1}=\frac{20-4\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*-1}=\frac{20+4\sqrt{5}}{-2}
$