-3(m-3)2m=12

Solution for -3(m-3)2m=12 equation:

-3(m-3)2m=12

We move all terms to the left:

-3(m-3)2m-(12)=0

We multiply parentheses

-6m^2+18m-12=0

a = -6; b = 18; c = -12;
Δ = b2-4ac
Δ = 182-4·(-6)·(-12)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-6}=\frac{-24}{-12}
=+2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-6}=\frac{-12}{-12}
=1
$