m^2=41

Solution for m^2=41 equation:

m^2=41

We move all terms to the left:

m^2-(41)=0

a = 1; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·1·(-41)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{41}}{2*1}=\frac{0-2\sqrt{41}}{2}
=-\frac{2\sqrt{41}}{2}
=-\sqrt{41}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{41}}{2*1}=\frac{0+2\sqrt{41}}{2}
=\frac{2\sqrt{41}}{2}
=\sqrt{41}
$