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