2(2m)(4m+2)=108

Solution for 2(2m)(4m+2)=108 equation:

2(2m)(4m+2)=108

We move all terms to the left:

2(2m)(4m+2)-(108)=0

We multiply parentheses

88m^2+44m-108=0

a = 88; b = 44; c = -108;
Δ = b2-4ac
Δ = 442-4·88·(-108)
Δ = 39952
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:

$\sqrt{\Delta}=\sqrt{39952}=\sqrt{16*2497}=\sqrt{16}*\sqrt{2497}=4\sqrt{2497}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{2497}}{2*88}=\frac{-44-4\sqrt{2497}}{176}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{2497}}{2*88}=\frac{-44+4\sqrt{2497}}{176}
$