p^2+9p-36-4p=0

Solution for p^2+9p-36-4p=0 equation:

p^2+9p-36-4p=0

We add all the numbers together, and all the variables

p^2+5p-36=0

a = 1; b = 5; c = -36;
Δ = b2-4ac
Δ = 52-4·1·(-36)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*1}=\frac{-18}{2}
=-9
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*1}=\frac{8}{2}
=4
$