$\lfloor (25x-2)/4 \rfloor =(13x+4)/3$ This problem is designated for those who have the basic knowledge of floor brackets 
I know the answer will be 6. But tell me how.

It is clear that $\dfrac{13x + 4}{3}$ must be an integer so $x$ must have the form of $2 + 3k$.
Substitute this into the equation and now we must solve:
$$\left\lfloor\frac{48+75k}{4}\right\rfloor = \left\lfloor 12 + \frac{75k}{4}\right\rfloor = 12 + \left\lfloor \frac{75k}{4}\right\rfloor = 13k + 10$$
$$ \left\lfloor \frac{75k}{4}\right\rfloor = 13k - 2$$
Find an approximate solution by solving $\frac{75k}{4} = 13k - 2$ and reaching $k = -\frac{8}{23}$.
We will now look for solutions on intervals around this point.
$$ \left.\left\lfloor \frac{75k}{4}\right\rfloor\right|_{k = -8/23} = -7$$
Let $13k -2 = -8,$ $k = -\dfrac{6}{13}.$
Let $13k -2 = -7,$ $k = -\dfrac{5}{13}.$
Let $13k -2 = -6,$ $k = -\dfrac{4}{13}.$
Let $13k -2 = -5,$ $k = -\dfrac{3}{13}.$
From these possible values of $k$ we find that $x = \dfrac{8}{13}, \dfrac{11}{13}, \dfrac{14}{13}, \dfrac{17}{13}$.
Plugging in to the original equation to verify we find two extraneous solutions and that $\boxed{x = \dfrac{14}{13}, \dfrac{17}{13}}$.
This method will not work well in all cases. Because $\frac{75}{4} = 18.75 \not\approx 13$ I assumed that there are a maximum of four solutions. In a case where we must solve $ \lfloor \frac{75k}{4}\rfloor = 19k - 2$ there are five solutions. In $ \lfloor \frac{75k}{4}\rfloor = 18.75k - 2$ there will be infinitely many.
