6p^2+6p-72=0

Solution for 6p^2+6p-72=0 equation:

6p^2+6p-72=0

a = 6; b = 6; c = -72;
Δ = b2-4ac
Δ = 62-4·6·(-72)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-42}{2*6}=\frac{-48}{12}
=-4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+42}{2*6}=\frac{36}{12}
=3
$