\( \Large x^{2} = 6 + \sqrt{6+\sqrt{6 + \sqrt{6} ...... \infty}} \)

So, \( \Large x^{2} = 6 + \sqrt{x^{2}} \)

=>\( \Large x^{2} = 6 + x \)

=>\( \Large x^{2} - x - 6 = 0 \)

=>\( \Large x^{2} + 2x - 3x - 6 = 0 \)

=>\( \Large x \left(x+2\right) - 3 \left(x+2\right) = 0 \)

=>\( \Large \left(x - 3\right) \left(x + 2\right) = 0 \)

Therefore, x = 3