Question

please answer with explanation

Answer 1

put
$y = \sqrt{6 + \sqrt{ 6 + \sqrt{6 + ...} } }$
i.e.
$y = \sqrt{6 + y} \\ {y}^{2} = 6 + y \\ {y}^{2} - y - 6 = 0 \\ (y - 3)(y + 2) = 0 \\ y = 3 \: or \: - 2$
Since minus value can not be permitted inside root, y =3
Hope this helps.