Solve the following equation:

$\dfrac{2}{3}x^2 - \dfrac{1}{3}x - 1 = 0$

Given, $\dfrac{2}{3}x^2 - \dfrac{1}{3}x - 1 = 0$.

$\Rightarrow \dfrac{2x^2 - x - 3}{3} = 0$

⇒ 2x² - x - 3 = 0

⇒ 2x² - 3x + 2x - 3 = 0

⇒ x(2x - 3) + 1(2x - 3) = 0

⇒ (x + 1)(2x - 3) = 0

⇒ x + 1 = 0 or 2x - 3 = 0

⇒ x = -1 or 2x = 3

⇒ x = -1 or x = $\dfrac{3}{2}$.

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