Multiply the entire equation by x to eliminate the fraction.

Since $x \neq 0$, we can multiply both sides by $x$ without changing the validity of the equation.

$$x \cdot (x) - 6 \cdot (x) + \frac{2}{x} \cdot (x) = 0 \cdot (x)$$ $$x^2 - 6x + 2 = 0$$

The equation is in the standard quadratic form $ax^2 + bx + c = 0$, where $a=1$, $b=-6$, and $c=2$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substitute the values of $a$, $b$, and $c$ into the formula:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(2)}}{2(1)}$$ $$x = \frac{6 \pm \sqrt{36 - 8}}{2}$$ $$x = \frac{6 \pm \sqrt{28}}{2}$$

Simplify $\sqrt{28}$:

$$\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}$$

Substitute this back into the expression for $x$:

$$x = \frac{6 \pm 2\sqrt{7}}{2}$$

Factor out 2 from the numerator:

$$x = \frac{2(3 \pm \sqrt{7})}{2}$$

Cancel out the common factor of 2:

$$x = 3 \pm \sqrt{7}$$

The two solutions are $x_1 = 3 + \sqrt{7}$ and $x_2 = 3 - \sqrt{7}$.
Since $\sqrt{7}$ is approximately 2.646, neither $3 + \sqrt{7}$ nor $3 - \sqrt{7}$ is equal to zero.
Both solutions are valid.

The final answers are $x = 3 + \sqrt{7}, 3 - \sqrt{7}$.