2p(p-4)+9p(p-4)+9(p-4)=0

Solution for 2p(p-4)+9p(p-4)+9(p-4)=0 equation:

2p(p-4)+9p(p-4)+9(p-4)=0

We multiply parentheses

2p^2+9p^2-8p-36p+9p-36=0

We add all the numbers together, and all the variables

11p^2-35p-36=0

a = 11; b = -35; c = -36;
Δ = b2-4ac
Δ = -352-4·11·(-36)
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-53}{2*11}=\frac{-18}{22}
=-9/11
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+53}{2*11}=\frac{88}{22}
=4
$