What is this? A floor function?

$4x^{2} - 40\lfloor x \rfloor + 51 = 0$

First denote that $\lfloor x \rfloor \in \mathbb{Z}$ for all real numbers $x$ . So, let's choose a certain $p \in \mathbb{Z}$ .

$4x^{2} - 40p + 51 = 0$

Then the solutions of this equation are

$x = \pm \sqrt{\frac{40p - 51}{4}}$

One trivial property of the floor function is: $\lfloor x \rfloor = p \Leftrightarrow p \leq x < p+1$

This means that it doesn't matter what the value of $p$ is, $x$ has to be in the range from $p$ (inclusive) to $p + 1$ (exclusive) then. So, let's find all of our certain $p \in \mathbb{Z}$ that satisfy this condition

$p \leq \pm \sqrt{\frac{40p - 51}{4}} < p+1$

Solving these inequalities gives us the following solutions:

Negativ-Sign-Case: No Solutions.

Positive-Sign-Case: $\frac{3}{2} \leq p < \frac{5}{2}$ or $\frac{11}{2} < p \leq \frac{17}{2}$

In our solution ranges, there are only $4$ integers (namely $p=2, p=6, p=7$ and $p=8$ ).

Because of that, there can also only be $\boxed{4}$ solutions for $x$ .