(2^(x+2) + 2^(3-x))*x >= 33x решите пежл

Решите задачу:

$(2^{x+2}+2^{3-x})x \geq 33x\\ (2^{x+2}+2^{3-x})x-33x \geq 0\\ x(2^{x+2}+2^{3-x}-33) \geq 0\\ x(2^{x+2}+2^{3-x}-33) = 0\\ x_1 = 0\\ 2^{x+2}+2^{3-x}-33 = 0\\ 2^x*2^2 + 2^3*2^{-x}-33=0\\ 2^x = t\\ 4t+\frac{8}{t}-33=0\\ 4t^2-33t+8=0\\ D = 1089-128 = 961, \sqrt{D} = 31\\ t_1=\frac{33-31}{8} = \frac{1}{4}\\\\ t_2=\frac{33+31}{8} = 8\\\\ 2^x=\frac{1}{4}\\ 2^x = 2^{-2}\\ x_2=-2\\ 2^x = 8\\ 2^x=2^3\\ x_3=3\\\\ x \in [-2;0] \cup [3; +\infty)$