\left|3x+6\right|\ < 12 equals
3x+6<12 and 3x+6>-12 so we have to solve for x in each inequality:
3x+6<12
Subtract 6 from both sides
3x-6+6<12-6
3x<6
Divide each side by 3
x<2
Now, the other one:
3x+6>-12
3x>-12-6
3x>-18
x>-6
So your answer is -6<x<2 or D
