Solve the following equations by using formula:

$\dfrac{2x + 5}{3x + 4} = \dfrac{x + 1}{x + 3}$

Given,

$\dfrac{2x + 5}{3x + 4} = \dfrac{x + 1}{x + 3} \\[1em] \Rightarrow (2x + 5)(x + 3) = (x + 1)(3x + 4) \\[1em] \Rightarrow (2x^2 + 6x + 5x + 15) = (3x^2 + 4x + 3x + 4) \\[1em] \Rightarrow 2x^2 - 3x^2 + 11x - 7x + 15 - 4 = 0 \\[1em] \Rightarrow -x^2 + 4x + 11 = 0 \\[1em] \Rightarrow x^2 - 4x - 11 = 0 \text{ ( On multiplying equation by -1) }$

Comparing it with ax² + bx + c = 0
a= 1, b = -4, c = -11

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

$\Rightarrow \dfrac{-(-4) ± \sqrt{(-4)^2 - 4 \times 1 \times -11 }}{2 \times 1} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{16 + 44}}{2} \\[1em] \Rightarrow \dfrac{4 + \sqrt{60}}{2} \text{ or } \dfrac{4 - \sqrt{60}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{15}}{2} \text{ or } \dfrac{4 - 2\sqrt{15}}{2} \\[1em] \Rightarrow 2 + \sqrt{15} \text{ or } 2 - \sqrt{15}$

Hence, roots of the equation are $2 + \sqrt{15} , 2 - \sqrt{15}$.