Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$ 
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$

My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor  $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor +\lfloor x+\dfrac23 \rfloor $$
$$\lfloor 4x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor $$
Now :
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0 $$
$$(\lfloor x \rfloor)^2+(\lfloor x \rfloor)(\lfloor x+\dfrac13 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac23 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac12 \rfloor)+(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac13 \rfloor)\\(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac23 \rfloor) +\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor=0$$
Now what ?

HINT: 
If $0<b < 1$ then $\lfloor x + b \rfloor = \lfloor x \rfloor$ if  $x-\lfloor 
 x\rfloor + b < 1$ or $\lfloor x \rfloor + 1$ otherwise
you might have to look at the different cases that arise
