I want to solve $\lfloor{x^2-5x\rfloor} = -6$. So far, I wrote
$$x^2-5x \ge -6\implies x^2-5x+6 \ge 0$$ and get $(x-2)(x-3) \ge 0$. Therefore, $x$ must satisfy $$x \le 2 \text{ or } x\ge 3.$$ However, I do not know what to do next. I also tried writing $x = k + \alpha$ for some integer $k=\lfloor x \rfloor$ and $\alpha \in [0,1)$. Then, plug $k+\alpha$ into the equation $x^2-5x$ but could not get something meaningful.
How can I solve this equation?
On
By the definition of the floor function, if $\lfloor x^2-5x\rfloor=-6$, then it is necessarily the case that $-6\leq x^2-5x<-5$. You can, of course, consider this inequality as two separate inequalities, namely $-6\leq x^2-5x$ (which you've already correctly solved), and $x^2-5x<-5$. The set of values of $x$ which solve your original inequality may then be obtained by taking the intersection of the solution sets to the two aforementioned inequalities. Do you reckon you can complete the question using this fact?
The solutions of the given equation are the solutions of the following system of inequalities $$\begin{cases} x^2-5 x<-5\\ x^2-5 x\geq -6\\ \end{cases} $$ That is $$\frac{1}{2} \left(5-\sqrt{5}\right)<x\leq 2\lor 3\leq x<\frac{1}{2} \left(\sqrt{5}+5\right)$$