To find the zeros of the function \( f(x) = \log_3(x - 1) + \log_3(2x + 3) \), we first set the function to zero:
\[ \log_3(x - 1) + \log_3(2x + 3) = 0. \]
Using the logarithmic property \(\log_b(a) + \log_b(c) = \log_b(ac)\), we can combine the logs:
\[ \log_3((x - 1)(2x + 3)) = 0. \]
Since \(\log_3(a) = 0\) implies \(a = 1\), we have:
\[ (x - 1)(2x + 3) = 1. \]
Expanding the left-hand side gives:
\[ 2x^2 + 3x - 2x - 3 = 1 \quad \Rightarrow \quad 2x^2 + x - 3 = 1. \]
Subtracting 1 from both sides leads to:
\[ 2x^2 + x - 4 = 0. \]
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 1, c = -4 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = 1^2 - 4(2)(-4) = 1 + 32 = 33. \]
Now we can find the roots:
\[ x = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4}. \]
Thus, the zeros of the function \(f(x)\) are:
\[ x = \frac{-1 + \sqrt{33}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{33}}{4}. \]
Therefore, the complete answer for the zeros of \( f(x) \) is:
\[ \boxed{x = \frac{-1 \pm \sqrt{33}}{4}}. \]