m^2=148

Solution for m^2=148 equation:

m^2=148

We move all terms to the left:

m^2-(148)=0

a = 1; b = 0; c = -148;
Δ = b2-4ac
Δ = 02-4·1·(-148)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{37}}{2*1}=\frac{0-4\sqrt{37}}{2}
=-\frac{4\sqrt{37}}{2}
=-2\sqrt{37}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{37}}{2*1}=\frac{0+4\sqrt{37}}{2}
=\frac{4\sqrt{37}}{2}
=2\sqrt{37}
$