Уравнение - 4^(-x+0,5) - 7*2^-x - 4 < 0

Решите задачу:

$2 \cdot 4^{-x}-7 \cdot 2^{-x}-4\ \textless \ 0 \\ 2^{-x} =t \\ \begin{cases} t\ \textgreater \ 0 \\2t^2-7t-4\ \textless \ 0 \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} t\ \textgreater \ 0 \\(2t+1)(t-4)\ \textless \ 0 \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} t\ \textgreater \ 0 \\ - \frac{1}{2}\ \textless \ t\ \textless \ 4 \end{cases} \\ \\ =\ \textgreater \ 0\ \textless \ t\ \textless \ 4 \\ \\ 2^{-x}\ \textless \ 4\\ -x\ \textless \ 2\\ x\ \textgreater \ -2 =\ \textgreater \ \boxed{x \in (-2; + \infty)}$