0, \\ 2t^2-5t+2 \geq 0, \\ 2t^2-5t+2 = 0, \\ D=9>0, \\ t_1=\frac{1}{2}, t_2=2, \\ a=2>0, \\ \left [ {{t \leq \frac{1}{2},} \atop {t \geq2;}} \right. \\ \left [ {{2^{x}\leq \frac{1}{2},} \atop {2^{x}\geq2;}} \right. \left [ {{2^{x}\leq2^{-1},} \atop {2^{x}\geq2^1;}} \right. \left [ {{x \leq -1,} \atop {x \geq 1;}} \right. \\ x\in(-\infty;-1]\cup[1;+\infty)." alt="2^{2x+1} -5\cdot2^{x} +2 \geq 0, \\ 2\cdot2^{2x}-5\cdot2^{x} +2 \geq 0, \\ 2^{x}=t, t>0, \\ 2t^2-5t+2 \geq 0, \\ 2t^2-5t+2 = 0, \\ D=9>0, \\ t_1=\frac{1}{2}, t_2=2, \\ a=2>0, \\ \left [ {{t \leq \frac{1}{2},} \atop {t \geq2;}} \right. \\ \left [ {{2^{x}\leq \frac{1}{2},} \atop {2^{x}\geq2;}} \right. \left [ {{2^{x}\leq2^{-1},} \atop {2^{x}\geq2^1;}} \right. \left [ {{x \leq -1,} \atop {x \geq 1;}} \right. \\ x\in(-\infty;-1]\cup[1;+\infty)." align="absmiddle" class="latex-formula">