Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$. 
Find the domain of the function $f(x)=\dfrac{64+x^2+x}{9x^2+27x+9}$.

I don't know how to start nor how to do it. Please help me!
 A: Hint: 
Solve the quadratic equation: $$9x^2+27x+9=0\tag{$\star$}$$ The roots of $\text{($\star$)}$ will be the $x$-values where the function $f(x)$ is undefined. (Since division by zero is undefined) You'll then conclude that the domain of the function $f(x)$ is that of all real numbers except the roots of $\text{($\star$)}$.
A: The domain of a function is the set of argument values for which the function is defined. 
We need to make $\displaystyle\frac{64+x^2+x}{9x^2+27x+9}$ defined 
$\displaystyle\implies 9x^2+27x+9\ne0\iff x^2+3x+1\ne0$
Now, the roots of $\displaystyle x^2+3x+1=0$ are $x=\frac{-3\pm\sqrt5}2$ which will make the given function undefined
So, the domain will be  $\left(-\infty, \infty\right)$ excluding those two points i.e.,
$\displaystyle\left(-\infty, \frac{-3-\sqrt5}2\right)\cup\left(\frac{-3-\sqrt5}2,\frac{-3+\sqrt5}2\right)\cup\left(\frac{-3+\sqrt5}2,\infty\right)$ 
Also observe that $\displaystyle\lim_{x\to\infty}\frac{64+x^2+x}{9x^2+27x+9}=\lim_{x\to\infty}\frac{\frac{64}{x^2}+1+\frac1x}{9+\frac{27}x+\frac9{x^2}}=\frac19$
