\(\left(2^x-32\right)\left(x^2-\dfrac{36}{25}\right)=0\)
=>\(\left[{}\begin{matrix}2^x-32=0\\x^2-\dfrac{36}{25}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2^x=32=2^5\\x^2=\dfrac{36}{25}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)