Решите иррациональное неравенство помогите пожалуйста

[latex](x^2-2x)-\sqrt{x^2-2x}-12\ \textless \ 0\\ \sqrt{x^2-2x} = t[/latex] [latex]\begin{cases} t \geq 0 \\ t^2-t-12 \ \textless \ 0\end{cases} \ \textless \ =\ \textgreater \ \begin{cases} t \geq 0 \\ (t+3)(t-4) \ \textless \ 0\end{cases} \ \textless \ =\ \textgreater \ \begin{cases} t \geq 0 \\ -3 \ \textless \ t \ \textless \ 4\end{cases}\\ \\ =\ \textgreater \ 0 \leq t \ \textless \ 4[/latex] [latex]0 \leq \sqrt{x^2-2x} \ \textless \ 4\\ 0 \leq x^2-2x \ \textless \ 16\\ \begin{cases} x^2-2x \geq 0 \\ x^2-2x -16\ \textless \ 0 \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} x(x-2) \geq 0 \\ (x-1+\sqrt{17})(x-1-\sqrt{17})\ \textless \ 0\end{cases} \ \textless \ =\ \textgreater \ [/latex] [latex]\begin{cases} x \in (-\infty; 0] \cup [2; +\infty) \\ x \in (1-\sqrt{17}; 1+\sqrt{17}) \end{cases} =\ \textgreater \ \boxed {x \in (1-\sqrt{17};0] \cup [2; 1+\sqrt{17}) }[/latex]