25p^2+20p-23=0

Solution for 25p^2+20p-23=0 equation:

25p^2+20p-23=0

a = 25; b = 20; c = -23;
Δ = b2-4ac
Δ = 202-4·25·(-23)
Δ = 2700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-30\sqrt{3}}{2*25}=\frac{-20-30\sqrt{3}}{50}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+30\sqrt{3}}{2*25}=\frac{-20+30\sqrt{3}}{50}
$