3 | x-3 | + 9= 24?

3|x-3| + 9 = 24
3|x-3| = 15
|x-3| = 5
x-3 = ±5
x = 3±5 = -2, 8

I have no clue what you did there but the way I'd do is this:
3 |x-3| + 9 = 24
3 |x-3| = 15
|x-3| = 5

Case-by-case analysis:
Case 1:
x - 3 = 5
x = 8

Case 2:
-(x - 3) = 5
x - 3 = -5
x = -2


It's not that strange that there are two solutions. The modulus effectively turns the straight line that 3(x-3)+9 would represent into a V-shaped function. By equating this with 24 you're looking for intersections with a parallel to the x-axis.
If you visualize the two graphs you'll see that you could either have
- one exact point of intersection if the tip of the V touches the parallel (then this tip would be at y=24)
- two points of intersection if the line intersects both "legs" of the V (that's the case here), or
- no point of intersection if the line runs 'beneath' the V

I x - 3 I = 5
x - 3 = - 5 , x - 3 = 5
x = - 2 , x = 8