solve for b:
take the square root of both sides:
[tex]\begin{gathered} \sqrt[]{4(3b+2)^2}=\sqrt[]{64} \\ \sqrt[]{4}\cdot\sqrt[]{(3b+2)^2}=\pm8 \\ 2(3b+2)=\pm8 \\ 6b+4=\pm8 \\ so\colon \\ b=\frac{\pm8-4}{6} \end{gathered}[/tex]therefore:
[tex]\begin{gathered} b=\frac{2}{3} \\ or \\ b=-2 \end{gathered}[/tex]