Find the interval on which $ x^{2} - \lfloor x \rfloor - 3 < 0 $ holds. On what interval does the equation $ x^{2} - \lfloor x \rfloor - 3 < 0 $ hold?
My attempt: I tried sketching the graph, but it’s a bit complicated.
Is there any other approach?

Probably graphing $y=x^2-3$ and $y=\lfloor x\rfloor$ and seeing where the first graph is below the second is the easiest way to solve this, but here is another method:
$\textbf{1)}$ Since $\lfloor x\rfloor \le x$,
$x^2-3<\lfloor x\rfloor\implies x^2-3<x\implies x^2-x-3<0\implies x^2-x+\frac{1}{4}<\frac{13}{4}\implies$
$\left(x-\frac{1}{2}\right)^2<\frac{13}{4}\implies|x-\frac{1}{2}|<\frac{\sqrt{13}}{2}<2\implies-\frac{3}{2}<x<\frac{5}{2}$.
$\textbf{2)}$ Since $x-1<\lfloor x\rfloor$,  the inequality will be satisfied if $x^2-3\le x-1$:
$x^2-3\le x-1 \iff x^2-x-2\le0\iff (x-2)(x+1)\le 0\iff -1\le x\le2$.
$\textbf{3)}$ a) If $-\frac{3}{2}<x<-1$, then $\lfloor x\rfloor=-2$, 
so $x^2-3<\lfloor x\rfloor \iff x^2<1\iff -1<x<1$; so there is no solution in this case.
b) If $2<x<\frac{5}{2}$, then $\lfloor x\rfloor=2$, so $x^2-3<\lfloor x\rfloor \iff x^2<5\iff-\sqrt{5}<x<\sqrt{5}$;
and therefore if $2<x<\sqrt{5}$, the inequality is satisfied.
Therefore the inequality is satisfied for $-1\le x<\sqrt{5}$.
