Solve for x :

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5, x$ ≠ $-3, \dfrac{1}{2}$

Given,

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5 \\[1em] \Rightarrow \text{ Taking } y = \dfrac{2x - 1}{x + 3} \text{ the equation becomes} \\[1em] \Rightarrow 2y - \dfrac{3}{y} = 5 \\[1em] \Rightarrow 2y^2 - 3 = 5y \\[1em] \Rightarrow 2y^2 - 5y - 3 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -5 , c = -3

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

we obtain:

$\Rightarrow \dfrac{-(-5) ± \sqrt{(-5)^2 - 4 \times 2 \times -3}}{2 \times 2} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{25 + 24}}{4} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{5 ± 7}{4} \\[1em] \Rightarrow \dfrac{5 + 7}{4} \text{ or } \dfrac{5 - 7}{4}\\[1em] \Rightarrow \dfrac{12}{4} \text{ or } \dfrac{-2}{4} \\[1em] \Rightarrow 3 \text{ or } -\dfrac{1}{2} \\[1em]$

But,

$y = \dfrac{2x - 1}{x + 3} \\[1em] \therefore 3 = \dfrac{2x - 1}{x + 3} \text{ or } -\dfrac{1}{2} = \dfrac{2x - 1}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 2x - 1 \text{ or } -(x + 3) = 2(2x - 1) \\[1em] \Rightarrow 3x + 9 = 2x - 1 \text{ or } -x - 3 = 4x - 2 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \text{ or } -x - 4x = -2 + 3 \\[1em] \Rightarrow x = -10 \text{ or } -5x = 1 \\[1em] \Rightarrow x = -10 \text{ or } x = -\dfrac{1}{5} \\[1em]$

Hence roots of the given equations are -10 , $-\dfrac{1}{5}$.