Solve for x: 3|x-3| 2=14

a. No solution

b. x=-1, x=8.3

c. x=0, x=7

d. x=-1, x=7

I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:

3|x-3|+2=14 |minus 2

3|x-3|=12 |divide 3

|x-3|=4

From absolute value definition you've got two ways:

x-3=4 or x-3=-4

x=7 or x=-1

And answer d) passes