Compute a division with integer and fractional part I have a problem that I don't know how to solve:
Compute $[\frac{\sqrt{7}}{frac(\sqrt{7})}]$ 
Here's what I've tried: $[\sqrt{7}]=2 \rightarrow frac(\sqrt{7}) = \sqrt{7}-2 \rightarrow [\frac{\sqrt{7}}{frac(\sqrt{7})}] =[\frac{\sqrt{7}}{\sqrt{7}-2}] = [\frac{\sqrt{7}(\sqrt{7}+2)}{7-4}] = [\frac{7+2\sqrt{7}}{3}]$
But from here I don't know what to do anymore.

$$x\le \frac{7+2\sqrt{7}}{3}<x+1$$
$$3x\le 7+2\sqrt{7}<3x+3$$
$$3x-7\le 2\sqrt{7}<3x-4$$
$$\frac{3x-7}{2}\le \sqrt{7}<\frac{3x-4}{2}$$
$$\frac{9x^2-42x+49}{4}\le 7<\frac{9x^2-24x+16}{4}$$
$$9x^2-42x+49\le 28<9x^2-24x+16$$
where $x\in N$. There are integer solution is $x=4$.
