Solve the following equation using quadratic formula:

$\dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3}$

$\Rightarrow \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 - 4x - x + 4 + (x^2 - 2x - 3x + 6)}{x^2 - 4x - 2x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 - 5x + 4 + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 10x + 10 }{x^2 - 6x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \\[1em] \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \\[1em] \Rightarrow 10x^2 - 60x + 80 - (6x^2 - 30x + 30 ) = 0 \\[1em] \Rightarrow 10x^2 - 60x + 80 - 6x^2 + 30x - 30 = 0 \\[1em] \Rightarrow 4x^2 - 30x + 50 = 0 \\[1em] \Rightarrow 2(2x^2 - 15x + 25) = 0 \\[1em] \Rightarrow 2x^2 - 15x + 25 = 0.$

Comparing equation 2x² - 15x + 25 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -15 and c = 25.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(25)}}{2(2)} \\[1em] = \dfrac{15 \pm \sqrt{225 - 200}}{4} \\[1em] = \dfrac{15 \pm \sqrt{25}}{4} \\[1em] = \dfrac{15 \pm 5}{4} \\[1em] = \dfrac{15 + 5}{4} \text{ or } \dfrac{15 - 5}{4} \\[1em] = \dfrac{20}{4} \text{ or } \dfrac{10}{4} \\[1em] = 5 \text{ or } \dfrac{5}{2}.$

Hence, x = $\Big{5, \dfrac{5}{2}\Big}$.