4m^2+8=104

Solution for 4m^2+8=104 equation:

4m^2+8=104

We move all terms to the left:

4m^2+8-(104)=0

We add all the numbers together, and all the variables

4m^2-96=0

a = 4; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·4·(-96)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*4}=\frac{0-16\sqrt{6}}{8}
=-\frac{16\sqrt{6}}{8}
=-2\sqrt{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*4}=\frac{0+16\sqrt{6}}{8}
=\frac{16\sqrt{6}}{8}
=2\sqrt{6}
$