$pt \Leftrightarrow 8\sqrt{3(x-1)}=-x^2+20x-24\Leftrightarrow \begin{cases}192(x-1)=x^4-40x^3+448x^2-960x+576 \\ 10-2\sqrt{19} \le x \le 10+2\sqrt{19} \end{cases}$$\Leftrightarrow \begin{cases}x^4-40x^3+448x^2-1152x+768=0 \\ 10-2\sqrt{19} \le x \le 10+2\sqrt{19} \end{cases}$
$\Leftrightarrow \begin{cases}(x^2-24x+48)(x^2-16x+16)=0 \\ 10-2\sqrt{19} \le x \le 10+2\sqrt{19} \end{cases}$
$\Leftrightarrow x=8+4\sqrt3 $ v $x=12-4\sqrt6$