3p^2+13p=16

Solution for 3p^2+13p=16 equation:

3p^2+13p=16

We move all terms to the left:

3p^2+13p-(16)=0

a = 3; b = 13; c = -16;
Δ = b2-4ac
Δ = 132-4·3·(-16)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*3}=\frac{-32}{6}
=-5+1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*3}=\frac{6}{6}
=1
$