Simple....
you have: [tex]3x= \sqrt{12-12x} [/tex]
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--->>>
[tex](3x)^{2} = \sqrt{12-12x} ^{2} [/tex]
Leaving you with....
[tex] 9x^{2} =12-12x[/tex]
Move the terms over and set it equal to zero.
-->>
[tex] 9x^{2}+12x-12=0 [/tex]
Factor out a 3...
[tex]3( 3x^{2} +4x-4=0)[/tex]
What multiplies to -12 and adds to 4?
6*-2=-12
6+-2=4
Leaving you with...
[tex]3(3x-2)(x+2)=0[/tex]
Remember you're solving for 0....
3x-2=0
3x-2=0
+2 +2
3x=2
[tex] \frac{3x}{3} = \frac{2}{3} [/tex]
x=[tex] \frac{2}{3} [/tex]
Then-->>
x+2=0
x+2=0
-2 -2
x=-2
Thus, your answer.
