k^2=121

Solution for k^2=121 equation:

k^2=121

We move all terms to the left:

k^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:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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