[tex]\begin{gathered} -2|5+2x|-6\le-20 \\ |x|=x,x\ge0 \\ |x|=-x,x<0 \\ for_{\text{ }}x\ge0 \\ -2(5+2x)-6\le-20 \\ -10-4x-6\le-20 \\ -4x\le-4 \\ x\ge\frac{-4}{-4} \\ x\ge1 \\ \end{gathered}[/tex][tex]\begin{gathered} for_{\text{ }}x<0 \\ -2(-5-2x)-6\le-20 \\ 10+4x-6\le-20 \\ 4x\le-20-4 \\ 4x\le-24 \\ x\le-\frac{24}{4} \\ x\le-6 \end{gathered}[/tex]
Therefore, the solutions are:
[tex]\begin{gathered} x\ge1 \\ or \\ x\le-6 \end{gathered}[/tex]
