Question 452437
I cannot figure out where I am going wrong on this problem. I distributed solving the parentheses first, but no matter what I do, I keep getting this problem wrong. 
3x+2[1-3(x+2)]=2x 
This is my work..
3x+2[1-3(x+2)]=2x
3x+2[1-3x-6]=2x
3x+2-6x-12=2x
---------------
-3x -10 = 2x
I get x= -10.
---------------
3x+2[1-3(x+2)]=2x
2[1-3(x+2)] = -x
2 - 3(x+2) = -x
- 3(x+2) = -x-2
3(x+2) = x+2
3x+6 = x+2
2x+4 = 0
x = -2