Solve the following equations by using formula:

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 , x$ ≠ $0, 2.$

Given,

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 \\[1em] \Rightarrow \dfrac{x - 2 + x }{x(x - 2)} = 3 \\[1em] \Rightarrow \dfrac{2x - 2}{x^2 - 2x} = 3 \\[1em] \Rightarrow 2x - 2 = 3(x^2 - 2x) \\[1em] \Rightarrow 2x - 2 = 3x^2 - 6x \\[1em] \Rightarrow 3x^2 - 6x - 2x + 2 = 0 \\[1em] \Rightarrow 3x^2 - 8x + 2 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -8 , c = 2

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \\[1em] \Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{64 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{40}}{6} \\[1em] \Rightarrow \dfrac{8 ± 2\sqrt{10}}{6} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{10}}{3} \\[1em] \Rightarrow \dfrac{4 + \sqrt{10}}{3} \text{ or } \dfrac{4 - \sqrt{10}}{3}$

Hence roots of the given equations are $\dfrac{4 + \sqrt{10}}{3} , \dfrac{4 - \sqrt{10}}{3}$.