-8r(r-5)=0

Solution for -8r(r-5)=0 equation:

-8r(r-5)=0

We multiply parentheses

-8r^2+40r=0

a = -8; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-8)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-8}=\frac{-80}{-16}
=+5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-8}=\frac{0}{-16}
=0
$