Answer: \(x=-1.58\) or \(x=0.58\).

We start with the main method. Use a substitution such as \(u=a^x\) to turn the exponential equation into a quadratic equation.

Start with

\(3(2^{2x+1})-11(2^x)+3=0.\)

Since \(2^{2x+1}=2(2^{2x})\),

\(3(2^{2x+1})=6(2^{2x}).\)

Let \(u=2^x\). Then \(2^{2x}=u^2\), so the equation becomes

\(6u^2-11u+3=0.\)

Factorise:

\((3u-1)(2u-3)=0.\)

Thus

\(u=\frac13\quad\text{or}\quad u=\frac32.\)

So

\(2^x=\frac13\quad\text{or}\quad 2^x=\frac32.\)

Taking logarithms,

\(x=\frac{\ln(1/3)}{\ln2}\quad\text{or}\quad x=\frac{\ln(3/2)}{\ln2}.\)

Therefore

\(x=-1.58\quad\text{or}\quad x=0.58\)

to 2 decimal places.

This completes the solution and gives the required result.