Решите неравенство: log1/3 (x-2)+log1/3 (12-x)>2

Решите задачу:

$log_{1/3}(x-2)+log_{1/3}(12-x)\ \textgreater \ 2\; ,\; \; \; ODZ:\; \left \{ {{x\ \textgreater \ 2} \atop {12\ \textgreater \ x}} \right. \; ,\; 2\ \textless \ x\ \textless \ 12\\\\log_{1/3}(x-2)(12-x)\ \textgreater \ log_{1/3}(1/3)^2\\\\-x^2+14x-24\ \textless \ 1/9\; |\cdot (-9)\\\\9x^2-126x+216\ \textgreater \ -1\\\\9x^2-126x+217\ \textgreater \ 0\\\\\frac{D}{4}=(\frac{126}{2})^2-9\cdot 217=2016=16\cdot 126\; ,\; \sqrt{\frac{D}{4}}=4\sqrt{126}\\\\x_1= \frac{63-4\sqrt{126}}{9} \approx 2,01\; ;\; x_2= \frac{63+4\sqrt{126}}{9} \approx 53,95\\\\x\in (-\infty , \frac{63-4\sqrt{126}}{9} )\cup ( \frac{63+4\sqrt{126}}{9} ,+\infty )$

$2\ \textless \ x\ \textless \ 12\quad \Rightarrow \\\\x\in (2;\; \frac{63-4\sqrt{126}}{9} )$