$\begin{array}{l}
2^{2 x}-3.2^{(x+2)}+32=0 \\
\Rightarrow\left(2^{x}\right)^{2}-3.2^{x} \cdot 2^{2}+32=0
\end{array}$

Let $2^{x}$ be $y$.

$\begin{array}{l}
\therefore y^{2}-12 y+32=0 \\
\Rightarrow y^{2}-8 y-4 y+32=0 \\
\Rightarrow y(y-8)-4(y-8)=0 \\
\Rightarrow(y-8)=0 \text { or }(y-4)=0 \\
\Rightarrow y=8 \text { or } y=4 \\
\therefore 2^{x}=8 \text { or } 2^{x}=4 \\
\Rightarrow 2^{x}=2^{3} \text { or } 2^{x}=2^{2} \\
\Rightarrow x-2 \text { or } 3
\end{array}$

Hence, 2 and 3 are the roots of the given equation.