Solve the following equation using quadratic formula:

$\dfrac{3}{4}$x² - x - 1 = 0

Comparing equation $\dfrac{3}{4}$x² - x - 1 = 0 with ax² + bx + c = 0, we get :

a = $\dfrac{3}{4}$, b = -1 and c = -1.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4\Big(\dfrac{3}{4}\Big)(-1)}}{2\Big(\dfrac{3}{4}\Big)} \\[1em] = \dfrac{1 \pm \sqrt{1 + 3}}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{1 \pm \sqrt{4}}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{1 \pm 2}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{2(1 \pm 2)}{3} \\[1em] = \dfrac{2 \pm 4}{3} \\[1em] = \dfrac{2 + 4}{3} \text{ or } \dfrac{2 - 4}{3} \\[1em] = \dfrac{6}{3} \text{ or } \dfrac{-2}{3} \\[1em] = 2 \text{ or } \dfrac{-2}{3}.$

Hence, $x = \Big{2, \dfrac{-2}{3}\Big}$.