(i) 4x² + 4√3x + 3 = 0

On comparing the given equation with ax² + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

By using quadratic formula, we get,

x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

⇒ x = -4√3±√48-48/8

⇒ x = -4√3±0/8

∴ x = -√3/2 or x = -√3/2

(ii) 2x² + x + 4 = 0

On comparing the given equation with ax² + bx + c = 0, we get,

a = 2, b = 1 and c = 4

By using quadratic formula, we get

x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

⇒ x = \(\frac{-1 \pm \sqrt{1-32}}{4}\)

⇒ x = \(\frac{-1 \pm \sqrt{1-32}}{4}\)

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.