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

Solution for 2(m+4)-4m(m+2)=0 equation:

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

We multiply parentheses

-4m^2+2m-8m+8=0

We add all the numbers together, and all the variables

-4m^2-6m+8=0

a = -4; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·(-4)·8
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{-(-6)-2\sqrt{41}}{2*-4}=\frac{6-2\sqrt{41}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{41}}{2*-4}=\frac{6+2\sqrt{41}}{-8}
$