p^2-4p=121

Solution for p^2-4p=121 equation:

p^2-4p=121

We move all terms to the left:

p^2-4p-(121)=0

a = 1; b = -4; c = -121;
Δ = b2-4ac
Δ = -42-4·1·(-121)
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10\sqrt{5}}{2*1}=\frac{4-10\sqrt{5}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10\sqrt{5}}{2*1}=\frac{4+10\sqrt{5}}{2}
$