Solve the following equation by factorisation:

$\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2$

Given,

$\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2 \\[1em] \Rightarrow \dfrac{8(2 - x) - 3(x + 3)}{(x + 3)(2 - x)} = 2 \\[1em] \Rightarrow 8(2 - x) - 3(x + 3) = 2(x + 3)(2 - x) \\[1em] \Rightarrow 16 - 8x - 3x - 9 = 2(2x - x^2 + 6 - 3x) \\[1em] \Rightarrow 7 - 11x = 2(- x^2 - x + 6) \\[1em] \Rightarrow 7 - 11x = -2x^2 - 2x + 12 \\[1em] \Rightarrow 7 - 11x + 2x^2 + 2x - 12 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] x = -\dfrac{1}{2} \text{ or } x = 5$

Hence, the roots of given equation are $-\dfrac{1}{2}$, 5.