Solve the following equation by factorisation:

$\dfrac{2}{3}x^2 - \dfrac{1}{3}x = 1$

Given,

$\dfrac{2}{3}x^2 - \dfrac{1}{3}x = 1 \\[1em] \Rightarrow \dfrac{2}{3}x^2 - \dfrac{1}{3}x - 1 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow \dfrac{2}{3}x^2 \times 3 - \dfrac{1}{3}x \times 3 - 1 \times 3 = 0 \times 3 \text{ (Multiplying the equation by 3) } \\[1em] \Rightarrow 2x^2 - x - 3 = 0 \\[1em] \Rightarrow 2x^2 - 3x + 2x - 3 = 0 \\[1em] \Rightarrow x(2x - 3) + 1(2x - 3) = 0 \\[1em] \Rightarrow (x + 1)(2x - 3) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } 2x - 3 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -1 \text{ or } x = \dfrac{3}{2}$

Hence, the roots of given equation are -1, $\dfrac{3}{2}$.