Range of $y = \frac{x^2-2x+5}{x^2+2x+5}$? How do I approach this problem? My book gives answer as $[\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. I tried forming an equation in $y$ and putting discriminant greater than or equal to zero but it didn't work. Would someone please help me?
I get $x^2 (y-1) + 2x (y+1) + (5y-5) =0$ and discriminant gives $2y^2 - y + 2 \leq 0$, which has complex roots.

Hint:
The derivative of $\dfrac{x^2-2x+5}{x^2+2x+5}$ is 
$\dfrac{4 (x^2 - 5)}{x^2+2x+5}$ and so the critical points are $\pm \sqrt 5$.
Consider also $\displaystyle\lim_{x\to\pm\infty}\dfrac{x^2-2x+5}{x^2+2x+5}=1$.
