p^2=3p+108

Solution for p^2=3p+108 equation:

p^2=3p+108

We move all terms to the left:

p^2-(3p+108)=0

We get rid of parentheses

p^2-3p-108=0

a = 1; b = -3; c = -108;
Δ = b2-4ac
Δ = -32-4·1·(-108)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*1}=\frac{-18}{2}
=-9
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*1}=\frac{24}{2}
=12
$