Lg(x-6)-0,5lg2=lg3+lgкорень квадратный из x-10

Решите задачу:

$lg(x-6)-0,5lg2=lg3+lg\sqrt{x-10}\; ,\\\\ ODZ:\; \; \left \{ {{x-6\ \textgreater \ 0} \atop {x-10\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 6} \atop {x\ \textgreater \ 10}} \right. \; ,\; x\ \textgreater \ 10\\\\lg\frac{x-6}{\sqrt 2}=lg(3\sqrt{x-10})\\\\\frac{x-6}{\sqrt 2}=3\sqrt{x-10}\\\\\sqrt{x-10}\cdot \sqrt2=\frac{x-6}{3}\\\\2(x-10)=\frac{x^2-12x+36}{9} \\\\18x-180=x^2-24x+72\\\\x^2-42x+252=0$

$D/4=21^2-252=189\ ,\\\\ x_1=21-\sqrt{189}=21-3\sqrt{21}\ \textless \ 10\; ,\; x_2=21+3\sqrt{21}\ \textgreater \ 10\\\\Otvet:\; \; x=21+3\sqrt{21}=\sqrt{21}\cdot (\sqrt{21}+3)\; .$