-4b^2+80b=0

Solution for -4b^2+80b=0 equation:

-4b^2+80b=0

a = -4; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·(-4)·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80}{2*-4}=\frac{-160}{-8}
=+20
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*-4}=\frac{0}{-8}
=0
$