[tex]\begin{gathered} |4x-6|\ge8 \\ |x|=x,if_{\text{ }}x\ge0 \\ |x|=-x,if_{\text{ }}x<0 \\ x\ge0 \\ 4x-6\ge8 \\ x\ge\frac{7}{2} \\ ---- \\ x<0 \\ -4x+6\ge8 \\ x\le-\frac{1}{2} \end{gathered}[/tex]
Therefore:
[tex](-\infty,-\frac{1}{2}\rbrack\cup\lbrack\frac{7}{2},\infty)[/tex]
