p^2=11p+26

Solution for p^2=11p+26 equation:

p^2=11p+26

We move all terms to the left:

p^2-(11p+26)=0

We get rid of parentheses

p^2-11p-26=0

a = 1; b = -11; c = -26;
Δ = b2-4ac
Δ = -112-4·1·(-26)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-15}{2*1}=\frac{-4}{2}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+15}{2*1}=\frac{26}{2}
=13
$