Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$ 
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$

I have reached: 
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$. 
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit: 
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it? 

$y = 1 +\frac{2x}{x^2 + x + 1} = 1 + \frac{2}{1+x+1/x}$,  $x\ne 0.$
A) Let $x\gt 0:$
1) $x+1/x \ge 2$
AM-GM: $x+1/x \ge 2\sqrt{x×1/x}$
$y_{max} = 1 + \frac{2}{1+2} = 5/3$.
B) Let $x \lt 0:$
2) $x+1/x \le -2$
$y_{min} = 1 + \frac{2}{1- 2}  = 1+ (-2) = -1.$
Since 
$y = \frac{x^2 + 3x + 1}{x^2+x+1}$ is continuous on $\mathbb{R}$, we find:
Range$ = [-1,5/3].$
