Алгебра: 4^(x+1)-5*2^(x)+1<0

Решите задачу:

$\displaystyle 4^{x+1}-5*2^{x}+1\ \textless \ 0 \\ 4*4^x-5*2^{x}+1\ \textless \ 0 \\ 4*(2^2)^x-5*2^x+1\ \textless \ 0 \\ 4*2^{2x}-5*2^x+1\ \textless \ 0 \\ 2^x=t, t\ \textgreater \ 0 \\ 4*t^2-5t+1\ \textless \ 0 \\ D=5^2-4*4=25-16=9 \\ t_1= \frac{5-3}{4*2}= \frac{1}{4} \\ \\ t_2= \frac{5+3}{2*4}=1 \\ 4(t-1)(t- \frac{1}{4})\ \textless \ 0 \\ t\in( \frac{1}{4}; 1) \\ \\ 2^x=t_1 \\ 2^x= \frac{1}{4} \\ 2^x=2^{-2} \\ x=-2 \\ \\ 2^x=t_2 \\ 2^x=1 \\ x=0 \\ \\ x\in(-2;0)$