22=a^2

Solution for 22=a^2 equation:

22=a^2

We move all terms to the left:

22-(a^2)=0

We add all the numbers together, and all the variables

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

The end solution:

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