Irrational equation $\sqrt{9-4x}=p-2x$ 
The equation
  $$\sqrt{9-4x}=p-2x$$
  has exactly 2 real and different solutions only if parameter $p$ belongs to which set?

So what I see here, to have the solutions be real in the first place,
$9-4x\ge0$
So $-4x\ge-9 \Rightarrow x\le{9\over4}$
$x\in\left(-\infty,{9\over4}\right]$
From here I quadrate both sides so I get 2 different possibilities:
1. For $p\ge2x$:
$9-4x=p^2-4x+4x^2\\9-4x-p^2+4x-4x^2=0\\-4x^2-p^2+9=0$
And for this to have 2 real and different solutions, $D>0$
$-4(-4)(-p^2+9)>0\\16(-p^2+9)>0\\-16p^2+144>0\\-16p^2>-144\\p^2<9\\|p|<3 \Rightarrow p\in(-3,3)$
2. For $p<2x:$
$9-4x=-p^2+4x-4x^2\\9-4x+p^2+4x^2-4x=0\\4x^2-8x+p^2+9=0$
Again $D>0$
$(-8)^2-4\cdot4(p^2+9)>0\\64-16p^2-144>0\\p^2<-5\\p\notin\Bbb R$
The problem here is I don't know how to use $x\in\left(-\infty,{9\over4}\right]$ in both cases, so I don't exactly know the solution to this problem. From the textbook I got this question from the result  is $p\in\left[{9\over2},5\right)$

As already pointed out in a comment, $(p-2x)^2 = p^2-4px+4x^2$, which results in
$$4x^2 - 4(p-1)x+(p^2-9) = 0,$$
or
$$x = \frac{(p-1)\pm\sqrt{10-2p}}{2}.$$
For two real distinct roots, $10-2p > 0 \implies \color{red}{p < 5}$.
Also, $p \ge 2x$ implies $$(p-1)\pm\sqrt{10-2p} \le p \implies \sqrt{10-2p}\le 1 \implies \color{red}{p \ge \frac{9}{2}}.$$
Note that
