Given:
$\begin{array}{l}
4^{(x+1)}+4^{(1-x)}=10 \\
\Rightarrow 4^{x} \cdot 4+4^{1} \cdot \frac{1}{4^{4}}=10
\end{array}$

Let $4^{x}$ be $y$.

$\begin{array}{l}
\therefore 4 y+\frac{4}{y}=10 \\
\Rightarrow 4 y^{2}-10 y+4=0 \\
\Rightarrow 4 y^{2}-8 y-2 y+4=0
\end{array}$

$\begin{array}{l}
\Rightarrow 4 y(y-2)-2(y-2)=0 \\
\Rightarrow y=2 \text { or } y=\frac{2}{4}=\frac{1}{2} \\
\Rightarrow 4^{x}=2 \text { or } \frac{1}{2} \\
\Rightarrow 4^{x}=2^{2 x}=2^{1} \text { or } 2^{2 x}=2^{-1} \\
\Rightarrow x=\frac{1}{2} \text { or } x=\frac{-1}{2}
\end{array}$

Hence, $\frac{1}{2}$ and $\frac{-1}{2}$ are roots of the given equation.