Корень из 5x-x^2 < x-2пожалуйста срочно решите

Решите задачу:

$\sqrt{5x-x^2}\ \textless \ x-2\; \; \Leftrightarrow \; \; \left \{ {{x-2\ \textgreater \ 0,\; 5x-x^2 \geq 0} \atop {5x-x^2\ \textless \ (x-2)^2}} \right. \\\\ \left \{ {{x-2\ \textgreater \ 0,\; x^2-5x \leq 0} \atop {5x-x^2\ \textless \ x^2-4x+4}} \right. \; \left \{ {{x\ \textgreater \ 2,\; x(x-5) \leq 0} \atop {2x^2-9x+4\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 2,\; 0 \leq x \leq 5} \atop {2(x-4)(x-\frac{1}{2})\ \textgreater \ 0}} \right. \; \left \{ {{x\in (2,5\, ]} \atop {x\in (-\infty ,\frac{1}{2})\cup (4,+\infty )}} \right. \\\\x\in (4,5\, ]$