First, express \(9^x\) in terms of base 3: \(9^x = (3^2)^x = (3^x)^2\).

Let \(u = 3^x\). Then the equation becomes \(u^2 = u + 12\).

Rearrange to form a quadratic equation: \(u^2 - u - 12 = 0\).

Factor the quadratic: \((u - 4)(u + 3) = 0\).

Thus, \(u = 4\) or \(u = -3\). Since \(u = 3^x\) and \(3^x > 0\), we have \(u = 4\).

Therefore, \(3^x = 4\).

Take the natural logarithm of both sides: \(x \ln 3 = \ln 4\).

Solve for \(x\): \(x = \frac{\ln 4}{\ln 3} \approx 1.26186\).

Round to two decimal places: \(x = 1.26\).