ОГЭ 21 номер !!!!!!!!!!!!

Решите задачу:

$\left \{ {{\frac{2-x}{2+(3-x)^2}\ \textgreater \ 0} \atop {6-9x \leq 31-4x}} \right. \; \left \{ {\frac{-(x-2)}{x^2-6x+11}\ \textgreater \ 0} \atop {5x \geq -22}} \right. \; \left \{ {{-(x-2)\ \textgreater \ 0} \atop {x \geq -\frac{22}{5}}} \right. \; \left \{ {{x\ \textless \ 2} \atop {x \geq -4,4}} \right. \\\\x\in [\, -4,4\; ;\; 2)\\\\P.S.\; \; x^2-6x+11=0\; \; \to \; \; D=36-44=-8\ \textless \ 0\; \; \to \\\\net\; dejstvitelnux\; kornej\; \; i\; \; x^2-6x+11\ \textgreater \ 0\; pri\; x\in (-\infty ,+\infty )$