Solve the following equation by factorisation:

$\dfrac{x + 2}{x + 3} = \dfrac{2x - 3}{3x - 7}$

Given,

$\dfrac{x + 2}{x + 3} = \dfrac{2x - 3}{3x - 7} \\[1em] \Rightarrow (x + 2) \times (3x - 7) = (2x - 3) \times (x + 3) \\[1em] \Rightarrow 3x^2 - 7x + 6x - 14 = 2x^2 + 6x - 3x - 9 \\[1em] \Rightarrow 3x^2 - x - 14 = 2x^2 + 3x - 9 \\[1em] \Rightarrow 3x^2 - 2x^2 - x - 3x - 14 + 9 = 0 \\[1em] \Rightarrow x^2 - 4x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - 5x + x - 5 = 0 \\[1em] \Rightarrow x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (x - 5)(x + 1) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[1em] x = 5 \text{ or } x = -1 \\[1em]$

Hence, the roots of given equation are -1, 5.