-24p^2+40p=0

Solution for -24p^2+40p=0 equation:

-24p^2+40p=0

a = -24; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-24)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-24}=\frac{-80}{-48}
=1+2/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-24}=\frac{0}{-48}
=0
$