Solve the following equation:

21x² = 4(2x + 1)

Given, 21x² = 4(2x + 1).

⇒ 21x² = 8x + 4

⇒ 21x² - 8x - 4 = 0

⇒ 21x² - 14x + 6x - 4 = 0

⇒ 7x(3x - 2) + 2(3x - 2) = 0

⇒ (7x + 2)(3x - 2) = 0

⇒ 7x + 2 = 0 or 3x - 2 = 0

⇒ 7x = -2 or 3x = 2

⇒ x = -$\dfrac{2}{7}$ or x = $\dfrac{2}{3}$.

Hence, roots of the given equations are $-\dfrac{2}{7}, \dfrac{2}{3}$.