Solve (using formula) the equation :

$\dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2\dfrac{4}{15}$.

Given, equation :

$\Rightarrow \dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2\dfrac{4}{15} \\[1em] \Rightarrow \dfrac{x^2 + (x + 1)^2}{x(x + 1)} = \dfrac{34}{15} \\[1em] \Rightarrow 15[x^2 + (x + 1)^2] = 34x(x + 1) \\[1em] \Rightarrow 15x^2 + 15(x + 1)^2 = 34x^2 + 34x \\[1em] \Rightarrow 15x^2 + 15(x^2 + 1 + 2x) = 34x^2 + 34x \\[1em] \Rightarrow 15x^2 + 15x^2 + 15 + 30x = 34x^2 + 34x \\[1em] \Rightarrow 34x^2 - 15x^2 - 15x^2 + 34x - 30x - 15 = 0 \\[1em] \Rightarrow 4x^2 + 4x - 15 = 0.$

Comparing above equation, with ax² + bx + c = 0, we get :

a = 4, b = 4 and c = -15.

By formula,

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

Substituting values we get :

$x = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times (-15)}}{2 \times 4} \\[1em] = \dfrac{-4 \pm \sqrt{16 + 240}}{8} \\[1em] = \dfrac{-4 \pm \sqrt{256}}{8} \\[1em] = \dfrac{-4 \pm 16}{8} \\[1em] = \dfrac{-4 + 16}{8}, \dfrac{-4 - 16}{8} \\[1em] = \dfrac{12}{8}, -\dfrac{20}{8} \\[1em] = \dfrac{3}{2}, -\dfrac{5}{2}$

Hence, x = $-\dfrac{5}{2}$, $\dfrac{3}{2}$