Solve this

$3^{(x+2)}+3^{-x}=10$

$3^{x} .9+\frac{1}{3^{x}}=10$

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

$\therefore 9 y+\frac{1}{y}=10$

$\Rightarrow 9 y^{2}+1=10 y$

$\Rightarrow 9 y^{2}-10 y+1=0$

$\Rightarrow(y-1)(9 y-1)=0$

$\Rightarrow y-1=0$ or $9 y-1=0$

$\Rightarrow y=1$ or $y=\frac{1}{9}$

$\Rightarrow 3^{x}=1$ or $3^{x}=\frac{1}{9}$

$\Rightarrow 3^{x}=3^{0}$ or $3^{x}=3^{-2}$

$\Rightarrow x=0$ or $x=-2$

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