To solve the equation \((x-3)(x+4)(x^2-9)=0\), we need to find the values of \(x\) that make the equation true. This involves setting each factor equal to zero and solving for \(x\).

First, let's rewrite the equation with the factors clearly identified:

\[
(x-3)(x+4)(x^2-9) = 0
\]

We can see that the equation is already factored, but we notice that \(x^2 - 9\) can be factored further. \(x^2 - 9\) is a difference of squares, which can be factored as follows:

\[
x^2 - 9 = (x - 3)(x + 3)
\]

So, the original equation becomes:

\[
(x-3)(x+4)(x-3)(x+3) = 0
\]

Now, we set each factor equal to zero:

1. \(x - 3 = 0\)
2. \(x + 4 = 0\)
3. \(x - 3 = 0\) (this is the same as the first factor, so it doesn't provide a new solution)
4. \(x + 3 = 0\)

Solving these equations, we get:

1. \(x - 3 = 0 \implies x = 3\)
2. \(x + 4 = 0 \implies x = -4\)
3. \(x + 3 = 0 \implies x = -3\)

So, the solutions to the equation \((x-3)(x+4)(x^2-9)=0\) are:

\[
x = 3, -4, -3
\]

These are the values of \(x\) that satisfy the given equation.