[tex]Solution, solve\:for\:x,\:3\left|x-3\right|+2=14\quad :\quad x=-1\quad \mathrm{or}\quad \:x=7[/tex]
[tex]Steps:[/tex]
[tex]\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}, 3\left|x-3\right|+2-2=14-2[/tex]
[tex]\mathrm{Simplify}, 3\left|x-3\right|=12[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}3, \frac{3\left|x-3\right|}{3}=\frac{12}{3}[/tex]
[tex]\mathrm{Simplify}, \left|x-3\right|=4[/tex]
[tex]|f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a, x-3=-4\quad \quad \mathrm{or}\quad \:\quad \:x-3=4[/tex]
[tex]x-3=-4\quad :\quad x=-1[/tex]
[tex]x-3=4\quad :\quad x=7[/tex]
[tex]\mathrm{Combine\:the\:ranges}, x=-1\quad \mathrm{or}\quad \:x=7[/tex]
The correct answer is B. x=-1, x=7
Hope this helps!!!
