b^2+20b=80

Solution for b^2+20b=80 equation:

b^2+20b=80

We move all terms to the left:

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

a = 1; b = 20; c = -80;
Δ = b2-4ac
Δ = 202-4·1·(-80)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{5}}{2*1}=\frac{-20-12\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{5}}{2*1}=\frac{-20+12\sqrt{5}}{2}
$