p2=10p+11

Solution for p2=10p+11 equation:

p2=10p+11

We move all terms to the left:

p2-(10p+11)=0

We add all the numbers together, and all the variables

p^2-(10p+11)=0

We get rid of parentheses

p^2-10p-11=0

a = 1; b = -10; c = -11;
Δ = b2-4ac
Δ = -102-4·1·(-11)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-12}{2*1}=\frac{-2}{2}
=-1
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+12}{2*1}=\frac{22}{2}
=11
$