m^2=1220

Solution for m^2=1220 equation:

m^2=1220

We move all terms to the left:

m^2-(1220)=0

a = 1; b = 0; c = -1220;
Δ = b2-4ac
Δ = 02-4·1·(-1220)
Δ = 4880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4880}=\sqrt{16*305}=\sqrt{16}*\sqrt{305}=4\sqrt{305}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{305}}{2*1}=\frac{0-4\sqrt{305}}{2}
=-\frac{4\sqrt{305}}{2}
=-2\sqrt{305}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{305}}{2*1}=\frac{0+4\sqrt{305}}{2}
=\frac{4\sqrt{305}}{2}
=2\sqrt{305}
$