To solve the equation \((x + 2)(3x - 1) + 1 - 9x^2 = 0\), we first need to expand the left-hand side.

1. Expand \((x + 2)(3x - 1)\):

\[
(x + 2)(3x - 1) = x(3x) + x(-1) + 2(3x) + 2(-1) = 3x^2 - x + 6x - 2 = 3x^2 + 5x - 2
\]

2. Now substitute this back into the original equation:

\[
3x^2 + 5x - 2 + 1 - 9x^2 = 0
\]

3. Combine like terms:

\[
(3x^2 - 9x^2) + 5x + (-2 + 1) = -6x^2 + 5x - 1 = 0
\]

4. Rearranging gives:

\[
-6x^2 + 5x - 1 = 0
\]

To make it simpler, we can multiply through by -1:

\[
6x^2 - 5x + 1 = 0
\]

5. Now we can use the quadratic formula to solve for \(x\):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 6\), \(b = -5\), and \(c = 1\):

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6}
\]

\[
x = \frac{5 \pm \sqrt{25 - 24}}{12}
\]

\[
x = \frac{5 \pm \sqrt{1}}{12}
\]

\[
x = \frac{5 \pm 1}{12}
\]

This gives us two potential solutions:

1. \(x = \frac{5 + 1}{12} = \frac{6}{12} = \frac{1}{2}\)

2. \(x = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}\)

Thus, the solutions for the equation \((x + 2)(3x - 1) + 1 - 9x^2 = 0\) are:

\[
x = \frac{1}{2} \quad \text{and} \quad x = \frac{1}{3}
\]