Answer:
x = 1 or x = -1
Step-by-step explanation:
[tex]2^{2x+1} -5(2^x)+2=0\\\\2^{2x}\cdot2^1 -5(2^x)+2=0\\\\2(2^x)^2 -5(2^x)+2=0[/tex]
Substitute: t = 2ˣ {t>0}
[tex]2t^2-5t+2=0\\\\t=\dfrac{5\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2}=\dfrac{5\pm\sqrt{25-16}}4=\dfrac{5\pm3}4\\\\ t=\dfrac84=2\qquad or\qquad t=\dfrac24=\dfrac12[/tex]
[tex]2^x=2\quad\iff\quad x=1\\\\2^x=\frac12\quad\iff\quad x=-1[/tex]
