Simple....
you have: \(3x= \sqrt{12-12x}\)
So...first you know you're trying to find x..
But, you have a square root on one side...to remove this you'd have to square both sides--->>>
\((3x)^{2} = \sqrt{12-12x} ^{2}\)
Leaving you with....
\(9x^{2} =12-12x\)
Move the terms over and set it equal to zero.
-->>
\(9x^{2}+12x-12=0\)
Factor out a 3...
\(3( 3x^{2} +4x-4=0)\)
What multiplies to -12 and adds to 4?
6*-2=-12
6+-2=4
Leaving you with...
\(3(3x-2)(x+2)=0\)
Remember you're solving for 0....
3x-2=0
3x-2=0
+2 +2
3x=2
\(\frac{3x}{3} = \frac{2}{3}\)
x=\(\frac{2}{3}\)
Then-->>
x+2=0
x+2=0
-2 -2
x=-2
Thus, your answer.
