0=11p-p^2-28

Solution for 0=11p-p^2-28 equation:

0=11p-p^2-28

We move all terms to the left:

0-(11p-p^2-28)=0

We add all the numbers together, and all the variables

-(11p-p^2-28)=0

We get rid of parentheses

p^2-11p+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$