Way 3:With $x\geq 0\Rightarrow \sqrt{6(x^2-2x+4)}=\sqrt{4x^2+2(x-3)^2+6}>2|x|=2x$
$\rightarrow $ Disequations $\Leftrightarrow 2x-4+2\sqrt{x}\geq \sqrt{6(x^2-2x+4)}$ $(1)$
$x=0$ is not a solution of disequation
$x>0\rightarrow (1)\Leftrightarrow 2(\sqrt{x}-\frac{2}{\sqrt{x}})+2\geq \sqrt{6(x-2+\frac{4}{x}})$ $(2)$
Set: $\sqrt{x}-\frac{2}{\sqrt{x}}=t\Rightarrow t^2=.............$
$(2)\Leftrightarrow \left\{ \begin{array}{l} 2t+2\geq 0\\ 4t^2+8t+4\geq 6t^2+12\end{array} \right.\Leftrightarrow ............\Leftrightarrow t=2$
$\Leftrightarrow x=4+2\sqrt{3}$
Conclude...................