$\begin{array}{l}
3^{(x+2)}+3^{-x}=10 \\
3^{x} .9+\frac{1}{3^{x}}=10
\end{array}$

Let $3^{x}$ be equal to $y$.

$\begin{array}{l}
\therefore 9 y+\frac{1}{y}=10 \\
\Rightarrow 9 y^{2}+1=10 y \\
\Rightarrow 9 y^{2}-10 y+1=0 \\
\Rightarrow(y-1)(9 y-1)=0 \\
\Rightarrow y-1=0 \text { or } 9 y-1=0 \\
\Rightarrow y=1 \text { or } y=\frac{1}{9} \\
\Rightarrow 3 x^{x}=1 \text { or } 3^{x}=\frac{1}{9} \\
\Rightarrow 3^{x}=3^{0} \text { or } 3^{x}=3^{-2} \\
\Rightarrow x=0 \text { or } x=-2
\end{array}$

Hence, 0 and $-2$ are the roots of the given equation.