b2+20b=-80

Solution for b2+20b=-80 equation:

b2+20b=-80

We move all terms to the left:

b2+20b-(-80)=0

We add all the numbers together, and all the variables

b^2+20b+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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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