k^2=22

Solution for k^2=22 equation:

k^2=22

We move all terms to the left:

k^2-(22)=0

a = 1; b = 0; c = -22;
Δ = b2-4ac
Δ = 02-4·1·(-22)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{22}}{2*1}=\frac{0-2\sqrt{22}}{2}
=-\frac{2\sqrt{22}}{2}
=-\sqrt{22}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{22}}{2*1}=\frac{0+2\sqrt{22}}{2}
=\frac{2\sqrt{22}}{2}
=\sqrt{22}
$