To solve the equation $x - 6 + \frac{2}{x} = 0$, with the condition $x \neq 0$: Step 1: 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. $$ x \cdot (x) - 6 \cdot (x) + \frac{2}{x} \cdot (x) = 0 \cdot (x) $$ $$ x^2 - 6x + 2 = 0 $$ Step 2: Use the quadratic formula to solve for $x$. 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: $$ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(2)}}{2(1)} $$ $$ x = \frac{6 \pm \sqrt{36 - 8}}{2} $$ $$ x = \frac{6 \pm \sqrt{28}}{2} $$ Step 3: Simplify the radical and the expression for $x$. 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: $$ x = 3 \pm \sqrt{7} $$ Step 4: Verify the solutions against the condition $x \neq 0$. 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 $\boxed{x = 3 + \sqrt{7}, 3 - \sqrt{7}}$.