3p^2+16=25p

Solution for 3p^2+16=25p equation:

3p^2+16=25p

We move all terms to the left:

3p^2+16-(25p)=0

a = 3; b = -25; c = +16;
Δ = b2-4ac
Δ = -252-4·3·16
Δ = 433
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{433}}{2*3}=\frac{25-\sqrt{433}}{6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{433}}{2*3}=\frac{25+\sqrt{433}}{6}
$