Solve the following equations by using formula:

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

Given,

$\dfrac{x + 1}{x + 3} = \dfrac{3x + 2}{2x + 3} \\[0.5em] \Rightarrow (x + 1)(2x + 3) = (3x + 2)(x + 3) \text{ On cross multiplication}\\[0.5em] \Rightarrow 2x^2 + 3x + 2x + 3 = 3x^2 + 9x + 2x + 6 \\[0.5em] \Rightarrow 2x^2 + 5x + 3 = 3x^2 + 11x + 6 \\[0.5em] \Rightarrow 2x^2 - 3x^2 + 5x - 11x + 3 - 6 = 0 \\[0.5em] \Rightarrow -x^2 - 6x - 3 = 0 \\[0.5em] \Rightarrow x^2 + 6x + 3 = 0 \text{ (Multiplying equation by -1) } \\[0.5em]$

The equation is x² + 6x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 6 , c = 3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-6 ± \sqrt{6^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-6 ±\sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + \sqrt{24}}{2} \text { or } \dfrac{-6 - \sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + 2\sqrt{6}}{2} \text{ or } \dfrac{-6 - 2\sqrt{6}}{2} \\[1em] -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}$

Hence roots of the given equations are $-3 + \sqrt{6} , -3 - \sqrt{6}$.