(i) To solve the quadratic equation \(z^2 + (2\sqrt{6})z + 8 = 0\), we use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\sqrt{6}\), and \(c = 8\).

Calculate the discriminant: \(b^2 - 4ac = (2\sqrt{6})^2 - 4 \times 1 \times 8 = 24 - 32 = -8\).

Since the discriminant is negative, the roots are complex: \(z = \frac{-2\sqrt{6} \pm \sqrt{-8}}{2}\).

\(\sqrt{-8} = \sqrt{8}i = 2\sqrt{2}i\).

Thus, the roots are \(z = \frac{-2\sqrt{6} \pm 2\sqrt{2}i}{2} = -\sqrt{6} \pm \sqrt{2}i\).

Therefore, the roots are \(-\sqrt{6} - \sqrt{2}i\) and \(-\sqrt{6} + \sqrt{2}i\).

(ii) On the Argand diagram, plot the points \(-\sqrt{6} - \sqrt{2}i\) and \(-\sqrt{6} + \sqrt{2}i\).

(iii) The distance \(OA = OB = \sqrt{(-\sqrt{6})^2 + (\pm \sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}\).

The angle \(AOB\) is the angle between the vectors \(OA\) and \(OB\), which are symmetric about the real axis. Therefore, \(\angle AOB = 60^\circ\).

(iv) Since \(OA = OB = AB = 2\sqrt{2}\), triangle \(AOB\) is equilateral.