25p^2-15p-28=0

Solution for 25p^2-15p-28=0 equation:

25p^2-15p-28=0

a = 25; b = -15; c = -28;
Δ = b2-4ac
Δ = -152-4·25·(-28)
Δ = 3025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3025}=55$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-55}{2*25}=\frac{-40}{50}
=-4/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+55}{2*25}=\frac{70}{50}
=1+2/5
$