p^2+12p-72=0

Solution for p^2+12p-72=0 equation:

p^2+12p-72=0

a = 1; b = 12; c = -72;
Δ = b2-4ac
Δ = 122-4·1·(-72)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{3}}{2*1}=\frac{-12-12\sqrt{3}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{3}}{2*1}=\frac{-12+12\sqrt{3}}{2}
$