121=p^2

Solution for 121=p^2 equation:

121=p^2

We move all terms to the left:

121-(p^2)=0

We add all the numbers together, and all the variables

-1p^2+121=0

a = -1; b = 0; c = +121;
Δ = b2-4ac
Δ = 02-4·(-1)·121
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22}{2*-1}=\frac{-22}{-2}
=+11
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22}{2*-1}=\frac{22}{-2}
=-11
$