b^2=122

Solution for b^2=122 equation:

b^2=122

We move all terms to the left:

b^2-(122)=0

a = 1; b = 0; c = -122;
Δ = b2-4ac
Δ = 02-4·1·(-122)
Δ = 488
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{488}=\sqrt{4*122}=\sqrt{4}*\sqrt{122}=2\sqrt{122}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{122}}{2*1}=\frac{0-2\sqrt{122}}{2}
=-\frac{2\sqrt{122}}{2}
=-\sqrt{122}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{122}}{2*1}=\frac{0+2\sqrt{122}}{2}
=\frac{2\sqrt{122}}{2}
=\sqrt{122}
$