Floor equation $\lfloor 3x-x^2 \rfloor = \lfloor x^2 + 1/2 \rfloor$ 
Solve the equation:
$$\left \lfloor 3x-x^2  \right \rfloor = \left \lfloor x^2 + 1/2 \right \rfloor$$

In the solution it writes

We notice that $x^{2}+\frac{1}{2}> 0$ therfore $\left \lfloor x^2 - 1/2 \right \rfloor \geq 0$ . From there $\left \lfloor 3x-x^2  \right \rfloor = n \geq 0$.

So far all of this I understand but than it writes:

But $3x-x^2 \leq \frac{9}{4}$ and $\left \lfloor 3x-x^2 \right \rfloor< 3$.

I dont understand this last part how did they get $3x-x^2 \leq \frac{9}{4}$ ?
 A: If we put $-x^2+3x$ into translated form we get:
$$-(x^2-3x)=-\left[\left(x-\frac 3 2\right)^2-\left(\frac32\right)^2\right]=-\left(x-\frac 3 2\right)^2+\frac94$$
We can now see that this is $x\mapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $\frac 3 2$ units right and $\frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $\left(\frac 3 2, \frac 9 4\right)$.) Hence, for all $x$, $-x^2+3x\le \frac 9 4$.
Edit
I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:

... therefore $\left\lfloor x^2+\frac 1 2\right\rfloor\ge0$. From there...

In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.
For example, investigating the floor 0 case:
$$\begin{align}
\left\lfloor x^2+\frac 1 2\right\rfloor = 0&\Rightarrow 0\le x^2+\frac 1 2<1\\
x^2+\frac 1 2\ge 0&\Rightarrow x\in\mathbb R\\
x^2+\frac 1 2 < 1&\Rightarrow
x\in\left]-\frac{\sqrt 2}{2},\frac{\sqrt2}{2}\right[\\\\
\left\lfloor-x^2+3x\right\rfloor=0&\Rightarrow0\le-x^2+3x<1\\
-x^2+3x\ge0&\Rightarrow x\in\left[0, 3\right]\\
-x^2+3x<1&\Rightarrow x\in\left]-\infty,\frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left]\frac{3+\sqrt5}{2},\infty\right[
\end{align}$$
Taking the intersection of all these sets gives us part of the solution to the original problem:
$$x\in\left[0, \frac{3-\sqrt 5}{2}\right[$$
Similarly, investigating the floor 1 case gives us that the equation holds for $x\in\left[\frac{\sqrt 2}{2}, 1\right[$, and the floor 2 case gives us that the equation holds for $x\in\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$.
The solution set to the problem is thus:
$$\left[0, \frac{3-\sqrt 5}{2}\right[\;\;\bigcup\;\;\left[\frac{\sqrt 2}{2}, 1\right[\;\;\bigcup\;\;\left[\frac{\sqrt6}{2},\frac{\sqrt{10}}{2}\right[$$
