Solve the following equation:

4x² - 2x + $\dfrac{1}{4}$ = 0

Given, 4x² - 2x + $\dfrac{1}{4}$ = 0

⇒ $\dfrac{16x^2 - 8x + 1}{4}$ = 0

⇒ 16x² - 8x + 1 = 0

⇒ 16x² - 4x - 4x + 1 = 0

⇒ 4x(4x - 1) - 1(4x - 1) = 0

⇒ (4x - 1)(4x - 1) = 0

⇒ (4x - 1)² = 0

⇒ 4x - 1 = 0

⇒ x = $\dfrac{1}{4}$.

Hence, roots of the given equations are $\dfrac{1}{4}, \dfrac{1}{4}$