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?

HINT
A little bit different approach from the other users.
To begin with, notice that
\begin{align*}
\frac{\sqrt{(x - 1)(x + 3)}}{x + 2} & = \frac{\sqrt{x^{2} + 2x - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x^{2} + 4x + 4) - (2x + 4) - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x + 2)^{2} - 2(x + 2) - 3}}{x + 2}\\\\
& = \frac{\sqrt{[(x + 2) - 1]^{2} - 4}}{x + 2}\\\\
& = \frac{\sqrt{(x + 1)^{2} - 4}}{(x + 1) + 1}
\end{align*}
Now we can make the change of variable $u = x + 1$ and study the range of the transformed expression:
\begin{align*}
y = \frac{\sqrt{u^{2} - 4}}{u + 1} \Longleftrightarrow
\begin{cases}
y(u + 1) = \sqrt{u^{2} - 4}\\\\
|u| \geq 2
\end{cases} \Longleftrightarrow
\begin{cases}
(1 - y^{2})u^{2} - 2y^{2}u - y^{2} - 4 = 0\\\\
|u| \geq 2
\end{cases}
\end{align*}
Can you take it from here?
