Решите неравенство (x - 8)² меньше √3(x-8)

[latex](x-8)^2\ \textless \ \sqrt3(x-8)\\\\t=x-8,\; \; t^2-\sqrt3t\ \textless \ 0\\\\t(t-\sqrt3)\ \textless \ 0\\\\+++(0)---(\sqrt3)+++\\\\t\in(0,\sqrt3)\; \; \to \; \; \left \{ {{t\ \textgreater \ 0} \atop {t\ \textless \ \sqrt3}} \right. \\\\a)\; x-8\ \textgreater \ 0\; \; \to \; \; x\ \textgreater \ 8\\\\b)\; \; x-8\ \textless \ \sqrt3\; \; \to \; \; x\ \textless \ 8+\sqrt3\\\\x\in (8,8+\sqrt3)[/latex]