m*(m-8)=0

Solution for m*(m-8)=0 equation:

m(m-8)=0

We multiply parentheses

m^2-8m=0

a = 1; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*1}=\frac{0}{2}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*1}=\frac{16}{2}
=8
$