Discuss the nature of the roots of the following equation without actually solving it:

2x² - $2\sqrt{6}x$ + 3 = 0

Comparing 2x² - $2\sqrt{6}x$ + 3 = 0 with ax² + bx + c = 0 we get,

a = 2, b = $-2\sqrt{6}$ and c = 3.

We know that,

Discriminant (D) = b² - 4ac = $(-2\sqrt{6})^2$ - 4 × 2 × 3

= (4 × 6) - 24 = 24 - 24 = 0;

Since, b is irrational and discriminant equals to zero.

Hence, the roots are irrational and equal.