Solve the following equations by using formula:

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

The given equation is $\dfrac{2}{x + 2} - \dfrac{1}{x + 1} = \dfrac{4}{x + 4} - \dfrac{3}{x + 3}$

$\Rightarrow \dfrac{2(x + 1) - (x + 2)}{(x + 2)(x + 1)} = \dfrac{4(x + 3) - 3(x + 4)}{(x + 4)(x + 3)} \\[1em] \Rightarrow \dfrac{2x + 2 - x - 2}{x^2 + x + 2x + 2} = \dfrac{4x + 12 - 3x - 12}{x^2 + 3x + 4x + 12} \\[1em] \Rightarrow \dfrac{x}{x^2 + 3x + 2} = \dfrac{x}{x^2 + 7x + 12} \\[1em] \Rightarrow x(x^2 + 7x + 12) = x(x^2 + 3x + 2) \\[1em] \Rightarrow (x^3 + 7x^2 + 12x) = (x^3 + 3x^2 + 2x) \\[1em] \Rightarrow x^3 + 7x^2 + 12x - x^3 - 3x^2 - 2x = 0 \\[1em] \Rightarrow 4x^2 + 10x = 0 \\[1em] \Rightarrow 2x(2x + 5) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } 2x + 5 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -\dfrac{5}{2}$

Hence, roots of the equation are $0 ,-\dfrac{5}{2}$.