Without solving, comment upon the nature of roots of the following equation :

x² + $2\sqrt{3}x - 9$ = 0

Comparing x² + $2\sqrt{3}$x - 9 = 0 with ax² + bx + c = 0 we get,

a = 1, b = $2\sqrt{3}$ and c = -9.

We know that,

Discriminant = D = b² - 4ac = $(2\sqrt{3})$² - 4(1)(-9)

= 12 + 36 = 48; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.