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? 
 A: $$y=1+\frac{2x}{x^2+x+1}$$
The function $y(x)$ is continuous and differentiable in $\Bbb R$. Moreover
$\lim_{x\to\pm\infty}y(x)=1$, so $y(x)$ is bounded. This means that the range of $y$ is an interval $I$.
Then
$$y'=\frac{2x^2+2x+2-4x^2-2x}{(x^2+x+1)^2}=\frac{2-2x^2}{(x^2+x+1)^2}$$
which vanishes at $x=\pm1$.
Compute $y(1)$ and $y(-1)$ to find the endpoints of $I$.
A: You have shown that $y$ is in the range of the function
$$y = \frac{x^2 + 3x + 1}{x + 1}$$
if $x$ is a real-valued root of the quadratic equation
$$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$
For $x$ to be a real-valued root of the quadratic equation, its discriminant must be nonnegative.  The discriminant is
\begin{align*}
\Delta & = b^2 - 4ac\\
       & = (y - 3)^2 - 4(y - 1)(y - 1)\\
       & = y^2 - 6y + 9 - 4(y^2 - 2y + 1)\\
       & = -3y^2 + 2y + 5 
\end{align*}
Hence, we require that 
\begin{align*}
-3y^2 + 2y + 5 & \geq 0\\
3y^2 - 2y - 5 & \leq 0\\
3y^2 + 3y - 5y - 5 & \leq 0\\
3y(y + 1) - 5(y + 1) & \leq 0\\
(3y - 5)(y + 1) & \leq 0
\end{align*}
which holds if $y \in [-1,5/3]$.  Therefore, the range of 
$$y = \frac{x^2 + 3x + 1}{x^2 + x + 1}$$
is 
$$\text{Ran} = \left[-1, \frac{5}{3}\right]$$
A: Points you should check are where $\dfrac{dy}{dx} = 0$ and limits of the range $\lim_{\pm\infty} \dfrac{dy}{dx}$
A: $y = 1 +\frac{2x}{x^2 + x + 1} = 1 + \frac{2}{1+x+1/x}$,  $x\ne 0.$
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].$
