Решить неравенство, где 2 - основание логарифма

Решите задачу:

$1+log_2(2-x)\ \textgreater \ log_2(x^2+3x+2)\; ,\\\\ODZ:\; \; \left \{ {{2-x\ \textgreater \ 0} \atop {x^2+3x+2\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 2} \atop {(x+2)(x+1)\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 2} \atop {x\ \textgreater \ -1\; ,\; x\ \textless \ -2}} \right. \\\\x\in (-\infty ,-2)\cup (-1,2)\\\\\\log_22+log_2(2-x)\ \textgreater \ log_2(x^2+3x+2)\\\\log_2(4-2x)\ \textgreater \ log_2(x^2+3x+2)\\\\4-2x\ \textgreater \ x^2+3x+2\\\\x^2+5x-2\ \textless \ 0\\\\D=25+8=33\; ,\; \; x_1=\frac{-5-\sqrt{33}}{2}\approx -5,37\; \; ;\; \; x_2=\frac{-5+\sqrt{33}}{2}\approx 0,37\\\\+++(-\frac{5+\sqrt{33}}{2})---(\frac{-5+\sqrt{33}}{2})+++$

$x\in (-\frac{5+\sqrt{33}}{2}\; ,\; \frac{-5+\sqrt{33}}{2})\\\\Otvet:\; \; (-\frac{5+\sqrt{33}}{2}\; ;\; -2)\cup (-1\; ;\; \frac{-5+\sqrt{33}}{2})$