How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$

$$\ x^2-19\lfloor x\rfloor+88=0 \tag{*}$$
Start by solving $x^2-19x+88=0$. This factors as $(x-8)(x-11)=0$ giving solutions $x\in\{8,11\}$. Since these are integers, they are solutions of (*) as well.
There are no more solutions to (*) for $8<x<9$ since 8 was an ordinary root and $x^2$ is increasing.
For $9\le x<10$, we have $\lfloor x \rfloor = 9$, so that (*) reduces to 
$$x^2-19\cdot9+88=0 \implies x^2=171-88=83 \implies x=\sqrt{83}$$
For $10\le x<11$, we have $\lfloor x \rfloor = 10$, so that (*) reduces to 
$$x^2-19\cdot10+88=0 \implies x^2=190-88=102 \implies x=\sqrt{102}$$
There are no more solutions to (*) for $11<x<12$ since 11 was an ordinary root and $x^2$ is increasing, no are there any for $x\ge12$.
So the solutions are:
$$x\in\{8,\sqrt{83},\sqrt{102},11\}$$

For all $x$, we have
$$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$
So for $x>0$ and a fixed $\lfloor x \rfloor$, since $x^2$ is increasing in $x$, the expression in (*) is minimised when $x=\lfloor x \rfloor$, i.e. when $x$ is an integer. But then for the factorisation $(x-8)(x-11)$ we have
$$0\le x<8 \implies (x-8)(x-11)>0$$ 
as both factors are negative. So the LHS in (*) is at least this large, so is always positive for $0\le x<8$, so there are no solutions with $0\le x<8$. 
It is trivial to see that there are no solutions for $x<0$, as all terms on the LHS in (*) are then positive.
