We have the following:
[tex]2|3-3x|-8\ge10[/tex]solving for x
[tex]\begin{gathered} 2|3-3x|-8\ge10 \\ 2|3-3x|-8+8\ge10+8 \\ \frac{2}{2}|3-3x|\ge\frac{18}{2} \\ |3-3x|\ge9 \\ 3-3x\ge9\rightarrow-3x\ge6\rightarrow x\leq\frac{6}{-3}\rightarrow x\leq-2 \\ 3-3x\leq-9\rightarrow-3x\leq-12\rightarrow x\ge\frac{-12}{-3}\rightarrow x\ge4 \end{gathered}[/tex]The answer is:
[tex](-\infty,-2\rbrack\cup\lbrack4,\infty)[/tex]