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

All we need is $\sqrt{6x^2+3x+\dfrac34}+x\ne0$ and of course  $6x^2+3x+\dfrac34\ge0$
$6x^2+3x+\dfrac34=6\left(x+\dfrac14\right)^2+\dfrac34-\dfrac6{16}>0$  for all real $x$
Now $\sqrt{6x^2+3x+\dfrac34}+x=0\iff x=-\sqrt{6x^2+3x+\dfrac34}\le0$
$\sqrt{6x^2+3x+\dfrac34}=-x$
Squaring we get $5x^2+3x+\dfrac34=0$ but $5x^2+3x+\dfrac34=5\left(x+\dfrac3{10}\right)^2+\dfrac34-5\cdot\dfrac9{100}>0$  for all real $x$
So, the denominator is non-zero real for all real $x$
