Integer part of $\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$ Find the value of the following infinite series:
$$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$
Now, my doubt is whether it's​ $2$ or it's​ $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it completely attains the value of $3$. 
I would to like to see the proof (formal) as well instead of just intuitions.

Let $$x= \sqrt[3]{24+\sqrt[3]{24+...}}$$ $$x^3 = 24+\sqrt[3]{24+\sqrt[3]{24+...}}$$
Notice $x^3 -24$ will be itself so $$x^3-24=x$$ Solving for $x$ yields $$3, \frac{3+\sqrt{23}i}{2}, \frac{3-\sqrt{23}i}{2}$$ but since we are concerning real solutions, 3 is the answer. It will be infinitely close to 3 amd reaches 3 so the answer is 3. ^^
Edit: That symbol is usually called "floor function". 
