If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$. 
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$
  where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.

Not sure if all these denotions are conventions that always mean what I've mentioned.
Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
 A: Since $$x-1\lt \lfloor x\rfloor \le x,$$
we have
$$\begin{align}x-1\lt \frac{4x^2+51}{40}\le x&\iff 4x^2-40x+91\gt0\ \text{and}\ 4x^2-40x+51\le 0\\&\iff 1.5\le x\lt 3.5\ \text{or}\ 6.5\lt x\le 8.5.\end{align}$$
1) When $1.5\le x\lt 2\Rightarrow \lfloor x\rfloor=1$, 
$$4x^2-40\times 1+51=0$$
does not have any real solution.
2) When $2\le x\lt 3\Rightarrow \lfloor x\rfloor=2$, 
$$4x^2-40\times 2+51=0\Rightarrow x=\pm \sqrt{29}/2\Rightarrow x=\sqrt{29}/2.$$
3) When $3\le x\lt 3.5\Rightarrow \lfloor x\rfloor=3$, 
$$4x^2-40\times 3+51=0\Rightarrow x=\pm \sqrt{69}/2.$$
But these don't satisfy $3\le x\lt 3.5.$
4) When $6.5\lt x\lt 7\Rightarrow \lfloor x\rfloor=6$, 
$$4x^2-40\times 6+51=0\Rightarrow x=\pm 3\sqrt{21}/2\Rightarrow x=3\sqrt{21}/2.$$
5) When $7\le x\lt 8\Rightarrow \lfloor x\rfloor=7$, 
$$4x^2-40\times 7+51=0\Rightarrow x=\pm \sqrt{229}/2\Rightarrow x=\sqrt{229}/2.$$
6) When $8\le x\le 8.5\Rightarrow \lfloor x\rfloor=8$, 
$$4x^2-40\times 8+51=0\Rightarrow x=\pm \sqrt{269}/2\Rightarrow x=\sqrt{269}/2.$$
Hence, the answer is 
$$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$
A: So I've decided to answer my own question.
Let $k\le x < k+1$ with $k\in\mathbb Z$. Then $$4x^2-40k+51=0$$
Now we need to find the solution $x$ in terms of $k$. That solution has to belong to the interval $[k\:;k+1)$. We can easily find out that $x$ is actually $$x=\pm\sqrt{10k-12.75}$$
As I said, the solution has to be in the interval, thus $$\pm\sqrt{10k-12.75}\ge k\tag{1}$$
We're only talking about real numbers, hence $$10k-12.75\ge0$$
Which shows that $$k\ge 1.275\tag{2}$$ $$\implies x>0\implies x=\sqrt{10k-12.75}$$
(I've thus removed the $\pm$ sign from the $x$)
Thus we'll only talk about positive numbers ($x,k$ are positive). This lets us square both sides of the inequality $(1)$ (remember we don't need the $\pm$ anymore): $$10k-12.75\ge k^2$$
Which shows that $k\in [1.5\:;8.5]\tag{3}$
We also know that $$\sqrt{10k-12.75}< k+1$$
And by doing the same we see that $k\in (-\infty\:;2.5)\cup(5.5\:;+\infty)\tag{4}$
$(2)(3)(4)\Rightarrow k=2, 6, 7, 8$.
We can substitute this to $x=\sqrt{10k-12.75}$ and get that $$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$
A: Note that when $i$ is odd then $3|(2^{i}-2)$ and hence $\lfloor \frac {2^{i}}{3} \rfloor = \lfloor \frac {2^{i}-2+2}{3} \rfloor = \lfloor \frac {2^{i}-2}{3} + \frac {2}{3} \rfloor = \frac {2^{i}-2}{3}$. 
Also when $j$ is even, then $3|(2^{j}-1)$ and hence $\lfloor \frac {2^{j}}{3} \rfloor = \lfloor \frac {2^{j}-1+1}{3} \rfloor = \lfloor \frac {2^{j}-1}{3} + \frac {1}{3} \rfloor = \frac {2^{j}-1}{3}$.
Therefore,
$$ \sum_{r=0}^{2010} \lfloor \frac {2^{r}}{3} \rfloor = \sum_i \lfloor \frac {2^{i}}{3} \rfloor + \sum_j \lfloor \frac {2^{j}}{3} \rfloor = \sum_i \frac {2^{i}-2}{3} + \sum_j  \frac {2^{j}-1}{3}$$
