b^2-20b+80=0

Solution for b^2-20b+80=0 equation:

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}
$