Many many solutions!

$\begin{aligned} 4x^2 - 40\lfloor x \rfloor + 51 & = 0 \\ \implies 4x^2 & = 40\lfloor x \rfloor - 51 \end{aligned}$

$\begin{aligned} \implies 4 \lfloor x \rfloor ^2 & \le 40\lfloor x \rfloor - 51 < 4(\lfloor x \rfloor + 1)^2 \end{aligned}$

$\implies \begin{cases} 4\lfloor x \rfloor^2 - 40 \lfloor x \rfloor + 51 \le 0 & \implies (2\lfloor x \rfloor-3)(2\lfloor x \rfloor-17) \le 0 & \implies 2 \le \lfloor x \rfloor \le 8 \\ 4\lfloor x \rfloor^2 - 32 \lfloor x \rfloor + 55 > 0 & \implies (2\lfloor x \rfloor-5)(2\lfloor x \rfloor-11) > 0 & \implies \lfloor x \rfloor \le 2 \cup \lfloor x \rfloor \ge 6 \end{cases}$

$\implies \lfloor x \rfloor = \{2, 6, 7, 8\}$ , $\boxed{4}$ solutions.