25p^2-16p=0

Solution for 25p^2-16p=0 equation:

25p^2-16p=0

a = 25; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·25·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*25}=\frac{0}{50}
=0
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*25}=\frac{32}{50}
=16/25
$