Infinite square roots

Let $y = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2...∞ } } } }$

Therefore, we can say that,

$y = \sqrt{2 + y }$

$y^{2} = 2 + y$

$y^{2} - y - 2 = 0$

$(y-2)(y+1) = 0$

$y = \boxed{2, -1}$

y can not be equal to -1 since square root can never be negative. So, (\ \boxed{y = 2} )

\sqrt{2+\sqrt{2+\sqrt{2\sqrt{2+\sqrt{2}+}}}}........be x then,you can write it as,\sqrt{2+x} x=\sqrt{2+x} x^{2}=2+x that implies x=2