-32m^2+24m=0

Solution for -32m^2+24m=0 equation:

-32m^2+24m=0

a = -32; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·(-32)·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*-32}=\frac{-48}{-64}
=3/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*-32}=\frac{0}{-64}
=0
$