We are given: \(\displaystyle{4}{\left({3}^{{2}}{x}+{1}\right)}+{17}{\left({3}^{{x}}\right)}={7}\)

Rewrite \(3^2x+1\) as \(3^2x\times 3\) so that: \(\displaystyle{4}{\left({3}^{{2}}{x}{\left({3}\right)}\right)}+{17}{\left({3}^{{x}}\right)}={7}\)

\(\displaystyle{12}{\left({3}^{{2}}{x}\right)}+{17}{\left({3}^{{x}}\right)}={7}\)

Let \(\displaystyle{u}={3}^{{x}}\) so that

\(\displaystyle{12}{u}^{{2}}+{17}{u}={7}\)

\(\displaystyle{12}{u}^{{2}}+{17}{u}-{7}={0}\)

Factor the left side: \((3u-1)(4u+7)=0\)

By zero-product property, \(\displaystyle{u}=\frac{{1}}{{3}},-{\left(\frac{{7}}{{4}}\right)}\)

Solve for \(\displaystyle{3}^{{x}}={u}\):

\(\displaystyle{3}^{{x}}=\frac{{1}}{{3}}\)

\(\displaystyle{3}^{{x}}=-{\left(\frac{{7}}{{4}}\right)}\)

\(\displaystyle{3}^{{x}}={3}^{{-{{1}}}}\) no solution as \(\displaystyle{3}^{{x}}{>}{0}\)

\(x=-1\)

So, the solution of the equation is: \(x=-1\)