20p^2+72p=-36

Solution for 20p^2+72p=-36 equation:

20p^2+72p=-36

We move all terms to the left:

20p^2+72p-(-36)=0

We add all the numbers together, and all the variables

20p^2+72p+36=0

a = 20; b = 72; c = +36;
Δ = b2-4ac
Δ = 722-4·20·36
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-48}{2*20}=\frac{-120}{40}
=-3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+48}{2*20}=\frac{-24}{40}
=-3/5
$