Let the square root of u be a + ib. Then, \\((a + ib)^2 = u = 10 - 4\sqrt{6}i \\).

Expanding the left side, we have \\(a^2 - b^2 + 2abi = 10 - 4\sqrt{6}i \\).

Equating real and imaginary parts, we get:

1. \\(a^2 - b^2 = 10 \\)

2. \\(2ab = -4\sqrt{6} \\)

From equation 2, \\(ab = -2\sqrt{6} \\).

Substitute \\(b = \frac{-2\sqrt{6}}{a} \\) into equation 1:

\\(a^2 - \left(\frac{-2\sqrt{6}}{a}\right)^2 = 10 \\)

\\(a^2 - \frac{24}{a^2} = 10 \\)

Multiply through by \\(a^2 \\) to clear the fraction:

\\(a^4 - 10a^2 - 24 = 0 \\)

Let \\(x = a^2 \\), then \\(x^2 - 10x - 24 = 0 \\).

Solving this quadratic equation using the quadratic formula:

\\(x = \frac{10 \pm \sqrt{100 + 96}}{2} = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2} \\)

Thus, \\(x = 12 \\) or \\(x = -2 \\) (discard \\(x = -2 \\) as it is not possible for \\(a^2 \\)).

So, \\(a^2 = 12 \\), giving \\(a = \pm \sqrt{12} = \pm 2\sqrt{3} \\).

Substitute back to find \\(b \\):

For \\(a = 2\sqrt{3} \\), \\(b = \frac{-2\sqrt{6}}{2\sqrt{3}} = -\sqrt{2} \\).

For \\(a = -2\sqrt{3} \\), \\(b = \frac{-2\sqrt{6}}{-2\sqrt{3}} = \sqrt{2} \\).

Thus, the square roots are \\(\pm (2\sqrt{3} - \sqrt{2}i) \\).