To solve the equation

\[
\frac{1}{x} + \frac{1}{x+2} = \frac{12}{35}
\]

we start by finding a common denominator for the left-hand side, which is \(x(x + 2)\). Rewriting both fractions, we have:

\[
\frac{x + 2}{x(x + 2)} + \frac{x}{x(x + 2)} = \frac{12}{35}
\]

Combining the fractions gives:

\[
\frac{(x + 2) + x}{x(x + 2)} = \frac{12}{35}
\]

This simplifies to:

\[
\frac{2x + 2}{x(x + 2)} = \frac{12}{35}
\]

Next, we can simplify \(2x + 2\) to \(2(x + 1)\):

\[
\frac{2(x + 1)}{x(x + 2)} = \frac{12}{35}
\]

To eliminate the fraction, we cross-multiply:

\[
2(x + 1) \cdot 35 = 12 \cdot x(x + 2)
\]

This simplifies to:

\[
70(x + 1) = 12x^2 + 24x
\]

Expanding the left side gives:

\[
70x + 70 = 12x^2 + 24x
\]

Bringing all terms to one side results in:

\[
12x^2 + 24x - 70x - 70 = 0
\]

This can be simplified to:

\[
12x^2 - 46x - 70 = 0
\]

Next, we divide the entire equation by 2 for simplification:

\[
6x^2 - 23x - 35 = 0
\]

Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -23\), and \(c = -35\):

Calculating the discriminant:

\[
b^2 - 4ac = (-23)^2 - 4(6)(-35) = 529 + 840 = 1369
\]

Since the discriminant is a perfect square:

\[
\sqrt{1369} = 37
\]

Now substituting back into the quadratic formula:

\[
x = \frac{-(-23) \pm 37}{2 \cdot 6} = \frac{23 \pm 37}{12}
\]

Calculating the two potential solutions:

1. For \(23 + 37\):

\[
x = \frac{60}{12} = 5
\]

2. For \(23 - 37\):

\[
x = \frac{-14}{12} = -\frac{7}{6}
\]

Thus, the solutions to the equation are:

\[
x = 5 \quad \text{and} \quad x = -\frac{7}{6}
\]