m^2=0.125

Solution for m^2=0.125 equation:

m^2=0.125

We move all terms to the left:

m^2-(0.125)=0

We add all the numbers together, and all the variables

m^2-0.125=0

a = 1; b = 0; c = -0.125;
Δ = b2-4ac
Δ = 02-4·1·(-0.125)
Δ = 0.5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.5}}{2*1}=\frac{0-\sqrt{0.5}}{2}
=-\frac{\sqrt{}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.5}}{2*1}=\frac{0+\sqrt{0.5}}{2}
=\frac{\sqrt{}}{2}
$