Question

what is? means answer​

Answer 1

Let,

$x=\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}\\\\x^2=6+\sqrt{6+\sqrt{6+\dots}}\\\\x^2-6=\sqrt{6+\sqrt{6+\dots}}$

Here RHS can be considered as x because,

$x=\sqrt{6+\sqrt{6+\dots}}$

So,

$x^2-6=x\\\\x^2-x-6=0\\\\x^2-3x+2x-6=0\\\\x(x-3)+2(x-3)=0\\\\(x-3)(x+2)=0\\\\\implies\ \mathbf{x=3}\quad;\quad \mathbf{x=-2}$

But is $x=-2$ logically true?!