Solve this

Given :

$2^{2 x}-3.2^{(x+2)}+32=0$

$\Rightarrow\left(2^{x}\right)^{2}-3 \cdot 2^{x} \cdot 2^{2}+32=0$

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

$\therefore y^{2}-12 y+32=0$

$\Rightarrow y^{2}-8 y-4 y+32=0$

$\Rightarrow y(y-8)-4(y-8)=0$

$\Rightarrow(y-8)=0$ or $(y-4)=0$

$\Rightarrow y=8$ or $y=4$

$\therefore 2^{x}=8$ or $2^{x}=4$

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

$\Rightarrow x=2$ or 3

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