Find the domain of $\frac{x}{\sqrt{6x^2+3x+3/4}+x}$. My attempt:
Let's assume that $\sqrt{6x^2+3x+\frac{3}{4}}+x$  $>$ 0$$
 \rightarrow\sqrt{6x^2+3x+\frac{3}{4}}  > -x
 \rightarrow {6x^2+3x+\frac{3}{4}}  > x^2\\
  \rightarrow {5x^2+3x+\frac{3}{4}}  > 0\\$$
  If x $<$ 0 then $x^2$ $>$ 0, and ${5x^2+3x+\frac{3}{4}}  > 0$ 
 the domain is $(-\infty,+\infty)\\$
 But, I'm not totally convinced. Any help is appreciated.
 Thank you

We have two cases to check.
First,  $$6x^2 + 3x + \frac{3}{4} \ge 0$$
It is $\ge 0$ for all $x$. 
Next, $$6x^2 + 3x + \frac{3}{4} - x\ne 0$$
$$6x^2 + 2x + \frac{3}{4} \ne 0$$
Solving we get complex values for $x$. Therefore, no real values exist for $x$.
Thus, you are correct. The domain is: $(-\infty, \infty)$
