[tex]4^x-6\times2^{x+1}+32=0\\\\\left(2^2\right)^x-6\times2^x\times2+32=0\\\\\left(2^x\right)^2-12\times2^x+32=0\\\\subtitute\ 2^x=t > 0\\\\t^2-12t+32=0\\\\t^2-4t-8t+32=0\\\\t(t-4)-8(t-4)=0\\\\(t-4)(t-8)=0\iff t-4=0\ or\ t-8=0\\\\t=4\ or\ t=8\\\\t=2^x,\ therefore:\\\\2^x=4\ or\ 2^x=8\\\\2^x=2^2\ or\ 2^x=2^3\\\\\boxed{x=2\ or\ x=3}[/tex]