25p^2+15p+2=0

Solution for 25p^2+15p+2=0 equation:

25p^2+15p+2=0

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

$\sqrt{\Delta}=\sqrt{25}=5$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*25}=\frac{-20}{50}
=-2/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*25}=\frac{-10}{50}
=-1/5
$