Is ma solution correct to this equation? $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3$$
My take: $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3\implies {x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=9$$
Thus: $$x+3=9\implies x=6$$
Did I do it right?

You got the right answer. This is a recursive question. Take: $$u=\sqrt{x+u}$$ We know this is true because, if you evaluate it, it will come out to be: $u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}$ expanded to $\infty$.
We are given what $u$ is. It is $3$ according to the first equation. Thus, plugging it back in: $$3=\sqrt{x+3}$$ $$x=6$$
