Solve the following equations by using formula:

$\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3} , x$ ≠ $0, - \dfrac{3}{2}$.

The given equation is $\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3}$

$\Rightarrow \dfrac{4 - 3x}{x} = \dfrac{5}{2x + 3} \text{ (On taking L.C.M.) } \\[1em] \Rightarrow (4 - 3x)(2x + 3) = 5x \text{ (On Cross multiplication) } \\[1em] \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \\[1em] \Rightarrow 6x^2 + 5x + 9x - 8x - 12 = 0 \\[1em] \Rightarrow 6x^2 + 6x - 12 = 0 \\[1em] \Rightarrow 6(x^2 + x - 2) = 0 \\[1em] x^2 + x - 2 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = 1, c = -2

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

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 1 \times -2}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 8}}{2} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{9}}{2} \text{ or } \dfrac{-1 - \sqrt{9}}{2} \\[1em] \Rightarrow \dfrac{-1 + 3}{2} \text{ or } \dfrac{-1 - 3}{2} \\[1em] \Rightarrow \dfrac{2}{2} \text{ or } \dfrac{-4}{2} \\[1em] 1 \text{ or } -2$

Hence, roots of the given equation are 1, -2.