If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$? 
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to? 

Attempt: 
$y=  \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that? 
The answer is: 

 $c= 4$


Hint. Note that
$$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}\quad\mbox{and}\quad \lim_{x\to \pm \infty}f(x)=1.$$
Moreover $f$ is bounded over the real line if and only if $x^2+2x+c=(x+1)^2+c-1>0$ that is $c>1$. In the bounded case, in order to determine the range of the continuous function $f$, consider the values of $f$ at $\pm \sqrt{c}$ (where the derivative $f'$ is zero):
$$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)}<1<f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)}.$$ 
