Solve the following equation:

$\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2$

Given, $\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2$.

Taking L.C.M. we get,

$\Rightarrow \dfrac{8(2 - x) - 3(x + 3)}{(x + 3)(2 - x)} = 2$

By cross multiplication we get,

⇒ 8(2 - x) - 3(x + 3) = 2(x + 3)(2 - x)

⇒ 16 - 8x - 3x - 9 = 2(2x - x² + 6 - 3x)

⇒ 7 - 11x = 2(6 - x - x²)

⇒ 7 - 11x = 12 - 2x - 2x²

⇒ 2x² - 11x + 2x + 7 - 12 = 0

⇒ 2x² - 9x - 5 = 0

⇒ 2x² - 10x + x - 5 = 0

⇒ 2x(x - 5) + 1(x - 5) = 0

⇒ (2x + 1)(x - 5) = 0

⇒ 2x + 1 = 0 or x - 5 = 0

⇒ 2x = -1 or x = 5

⇒ x = $-\dfrac{1}{2}$ or x = 5.

Hence, roots of the given equations are $-\dfrac{1}{2}$, 5.