m^2=57

Solution for m^2=57 equation:

m^2=57

We move all terms to the left:

m^2-(57)=0

a = 1; b = 0; c = -57;
Δ = b2-4ac
Δ = 02-4·1·(-57)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{57}}{2*1}=\frac{0-2\sqrt{57}}{2}
=-\frac{2\sqrt{57}}{2}
=-\sqrt{57}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{57}}{2*1}=\frac{0+2\sqrt{57}}{2}
=\frac{2\sqrt{57}}{2}
=\sqrt{57}
$