p+p(p-8)+3p=112

Solution for p+p(p-8)+3p=112 equation:

p+p(p-8)+3p=112

We move all terms to the left:

p+p(p-8)+3p-(112)=0

We add all the numbers together, and all the variables

4p+p(p-8)-112=0

We multiply parentheses

p^2+4p-8p-112=0

We add all the numbers together, and all the variables

p^2-4p-112=0

a = 1; b = -4; c = -112;
Δ = b2-4ac
Δ = -42-4·1·(-112)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{29}}{2*1}=\frac{4-4\sqrt{29}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{29}}{2*1}=\frac{4+4\sqrt{29}}{2}
$