Bardzo proszę o rozwiązanie zadania z załącznika!

[tex]2^{x^{2} -2} \cdot 4^{x+3} \leq 8^{x+2} \\\\2^{x^{2} -2} \cdot (2^{2} )^{x+3} \leq (2^{3} )^{x+2} \\\\2^{x^{2} -2} \cdot 2^{2\cdot (x+3)} \leq 2^{3\cdot (x+2)}\\\\2^{x^{2} -2} \cdot 2^{2x+6} \leq 2^{3x+6}\\\\2^{x^{2} -2+2x+6} \leq 2^{3x+6}\\\\2^{x^{2}+2x+4} \leq 2^{3x+6}\\\\~~~~\Downarrow \\\\x^{2}+2x+4\leq 3x+6\\\\x^{2} -x-2 \leq 0\\\\\Delta = (-1)^{2} -4\cdot 1\cdot (-2)\\\\\Delta = 1+8\\\\\Delta = 9~~,~~\sqrt{\Delta}=3\\\\[/tex]
[tex]x_{1} =\dfrac{1-3}{2} ~~~\lor ~~x_{2} =\dfrac{1+3}{2} \\\\x_{1} =\dfrac{-2}{2} ~~~\lor ~~x_{2} =\dfrac{4}{2} \\\\x_{1} =-1~~\lor~~x_{2}=2\\\\x^{2} -x-2\leq 0\\\\a=1~~~\Rightarrow ~~ramiona ~~funkcji~~kwadratowej~~skierowane ~~do~~gory \\\\x\in (-\infty ; -1 > \cup <2 ; + \infty )\\\\Odp: ~~2^{x^{2} -2} \cdot 4^{x+3} \leq 8^{x+2} ~~\Leftrightarrow ~~x\in (-\infty ; -1 > \cup <2 ; + \infty )[/tex]