To solve the equation \( x^2 - 9 - (x - 3)(2x - 5) = 0 \), let's start by expanding the term \((x - 3)(2x - 5)\).

1. **Expand \((x - 3)(2x - 5)\)**:

\[
(x - 3)(2x - 5) = x(2x - 5) - 3(2x - 5) = 2x^2 - 5x - 6x + 15 = 2x^2 - 11x + 15
\]

2. **Substitute this back into the equation**:

\[
x^2 - 9 - (2x^2 - 11x + 15) = 0
\]

This simplifies to:

\[
x^2 - 9 - 2x^2 + 11x - 15 = 0
\]

3. **Combine like terms**:

\[
-x^2 + 11x - 24 = 0
\]

To make it easier to work with, multiply the whole equation by -1:

\[
x^2 - 11x + 24 = 0
\]

4. **Now, we can factor this quadratic equation**:

We want two numbers that multiply to \(24\) and add to \(-11\). These numbers are \(-3\) and \(-8\):

\[
(x - 3)(x - 8) = 0
\]

5. **Set each factor equal to zero**:

\[
x - 3 = 0 \quad \text{or} \quad x - 8 = 0
\]

Thus, we find:

\[
x = 3 \quad \text{or} \quad x = 8
\]

6. **Conclusion**:

The solutions for the equation \( x^2 - 9 - (x - 3)(2x - 5) = 0 \) are:

\[
\boxed{3 \text{ and } 8}
\]