The given equation is $9x^2 + 2(3+2)x + 1 = 0$. We need to solve for $x$.

1. Problem Description

The given equation is $9x^2 + 2(3+2)x + 1 = 0$. We need to solve for $x$.

2. Solution Steps

First, simplify the equation. Since $3+2 = 5$, the equation becomes:

$9x^2 + 2(5)x + 1 = 0$

$9x^2 + 10x + 1 = 0$

Now, we can use the quadratic formula to solve for $x$. The quadratic formula is:

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

In our equation, $a = 9$, $b = 10$, and $c = 1$. Plugging these values into the quadratic formula, we get:

$x = \frac{-10 \pm \sqrt{10^2 - 4(9)(1)}}{2(9)}$

$x = \frac{-10 \pm \sqrt{100 - 36}}{18}$

$x = \frac{-10 \pm \sqrt{64}}{18}$

$x = \frac{-10 \pm 8}{18}$

So, we have two possible solutions for $x$:

$x_1 = \frac{-10 + 8}{18} = \frac{-2}{18} = -\frac{1}{9}$

$x_2 = \frac{-10 - 8}{18} = \frac{-18}{18} = -1$

3. Final Answer

The solutions are $x = -\frac{1}{9}$ and $x = -1$.