<s>[INST] 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.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
 [/INST] Let $x = n + \epsilon$ where $n\in \mathbb Z$ and $0 \le \epsilon \lt 1$.
Then
$x^2 = (n + \epsilon)^2 = 19n - 88$
Since $0 \le \epsilon \lt 1$, then  $n^2 \le 19n - 88 \lt (n+1)^2$
The line $y = 19x - 88$ intersects the parabola $y = x^2$ at $x = 8$ and at $x = 11$. Since the parabola $y = x^2$ is concave up, this implies that
$n \in \{8, 9, 10, 11 \}$ and $x = \sqrt{19n - 88}$
$n = 8 \implies x = 8$
$n = 9 \implies x = \sqrt{83}$
$n = 10 \implies x = \sqrt{102}$
$n = 11 \implies x = 11$
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