-4p+p^2=-132p-5p^2=7

Solution for -4p+p^2=-132p-5p^2=7 equation:

-4p+p^2=-132p-5p^2=7

We move all terms to the left:

-4p+p^2-(-132p-5p^2)=0

We get rid of parentheses

p^2+5p^2+132p-4p=0

We add all the numbers together, and all the variables

6p^2+128p=0

a = 6; b = 128; c = 0;
Δ = b2-4ac
Δ = 1282-4·6·0
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-128}{2*6}=\frac{-256}{12}
=-21+1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+128}{2*6}=\frac{0}{12}
=0
$