Square roots- I don't know what to do.

If $\sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=9$, find x.

I know that if this equation were switched around, where the x is 9 and 9 is x instead, we'd solve it like this:

$\sqrt{9+\!\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}}=x\\ x^2=9+\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}\\ x^2=9+x\\ x^2-x-9=0\\ x=\frac{1+\sqrt{37}}{2}$

I think I should be applying the same logic, but it's kind of difficult because of the x and it's making me confused. Could anyone help?