Find the roots of the following equation:
$\frac{1}{x} - \frac{1}{x-2} = 3, x \neq 0, 2$

(N/A) Given the equation: $\frac{1}{x} - \frac{1}{x-2} = 3$ for $x \neq 0, 2$.
Multiplying both sides by $x(x-2)$,we get:
$(x-2) - x = 3x(x-2)$
Simplifying the left side:
$-2 = 3x^2 - 6x$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$3x^2 - 6x + 2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 3, b = -6, c = 2$:
Discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$.
Thus,$x = \frac{6 \pm \sqrt{12}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6}$.
Dividing by $2$:
$x = \frac{3 \pm \sqrt{3}}{3}$.
Therefore,the roots are $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$.