Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$ 
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$

My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies  x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?

First notice that
$${\rm Dom}(f)=\left]-\infty,-3\right]\cup \left[1,+\infty\right[$$
Now, setting
$$y=f(x)\implies y=\frac{\sqrt{(x-1)(x+3)}}{x+2}\implies y^{2}=\frac{(x-1)(x+3)}{(x+2)^{2}}$$
Then,
$$y^{2}(x+2)^{2}=(x-1)(x+3)$$
Expanding and collecting all in terms of $x$, we have
$$(1-y^{2})x^{2}+(2-4y^{2})x+(-3-4y^{2})=0,\quad (*)$$
Using the quadratic formula we get
$$x=\frac{1-2y^{2}\pm \sqrt{4-3y^{2}}}{y^{2}-1},\quad (**)$$
Then,

*

*If $y=-1$, then $x=-\frac{7}{2}\in {\rm Dom}(f)$ using $(*)$ so not problem.


*If $y\not=\pm 1$ so only we need $4-3y^{2}\geqslant 0$ so $-\frac{2}{\sqrt{3}}\leqslant y\leqslant \frac{2}{\sqrt{3}}$. $[◇]$
Joint all, we have
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},1\right[$$
as desired.
Fixed and more details:
Notice with $y=1$, we have $1=\frac{\sqrt{(x-1)(x+3)}}{x+2}$  which don't have solutions over $\mathbb{R}$. Then exclude the $1$ of the interval.
Well, by $(**)$ we get
$$y\in \left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[\cup \left]-1,\frac{2}{\sqrt{3}}\right[$$
then we have the conclusion give in $[◇]$. However the interval $\left]-1,\frac{2}{\sqrt{3}}\right[$ is a strange solution (just study $f$ using the domain and the conclusion as follows), so we remove it. In my previous post I included the value $1$ in the range, however it was a mistake because as said above. Therefore the range of $f$ over the natural domain is given by
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[=\left[-\frac{2}{\sqrt{3}},1\right[.$$
