The expression is determined with all \(x\in R\)

\(\Leftrightarrow3\left(x^2+x+3\right)-\sqrt{x^2+x+3}-24=0\)

Let \(\sqrt{x^2+x+3}=a>0\Rightarrow3a^2-a-24=0\Leftrightarrow\left(a-3\right)\left(3a+8\right)=0\)

\(\Leftrightarrow a=3\left(cause:3a+8>0\right)\)

So  the expression is equal : \(\sqrt{x^2+x+3}=3\)

\(\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

So the expression has two solutions : x = 2 ; x = -3.