Solve the following equation using quadratic formula:

$\dfrac{x + 3}{2x + 3} = \dfrac{x + 1}{3x + 2}$

$\Rightarrow \dfrac{x + 3}{2x + 3} = \dfrac{x + 1}{3x + 2} \\[1em] \Rightarrow (x + 3)(3x + 2) = (x + 1)(2x + 3) \\[1em] \Rightarrow (3x^2 + 2x + 9x + 6) = (2x^2 + 3x + 2x + 3) \\[1em] \Rightarrow 3x^2 + 11x + 6 = 2x^2 + 5x + 3 \\[1em] \Rightarrow 3x^2 - 2x^2 + 11x - 5x + 6 - 3 = 0 \\[1em] \Rightarrow x^2 + 6x + 3 = 0.$

Comparing equation x² + 6x + 3 = 0 with ax² + bx + c = 0, we get :

a = 1, b = 6 and c = 3.

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(6) \pm \sqrt{(6)^2 - 4 \times (1) \times (3)}}{2 \times 1} \\[1em] = \dfrac{-6 \pm \sqrt{36 - 12}}{2} \\[1em] = \dfrac{-6 \pm \sqrt{24}}{2} \\[1em] = \dfrac{-6 \pm \sqrt{6 \times 4}}{2} \\[1em] = \dfrac{-6 \pm 2\sqrt{6}}{2} \\[1em] = \dfrac{2(-3 \pm \sqrt{6})}{2} \\[1em] = -3 \pm \sqrt{6} \\[1em] = -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}.$

Hence, $x = \Big{(-3 + \sqrt{6}), (-3 - \sqrt{6})\Big}$.