2. Solve the inequality $8|x-\sqrt{x}+2|+2 x \sqrt{x}<x^{2}+x+28$.
Answer: $x \in[0 ; 4) \cup(9 ;+\infty)$.

Solution. Note that the expression under the modulus is non-negative on the domain of definition (it is a quadratic trinomial in terms of $\sqrt{x}$ and $D0$, from which $t \in(-\infty ; 2) \cup(6 ;+\infty)$.

If $t6$, then $x-\sqrt{x}-6>0 \Leftrightarrow(\sqrt{x}-3)(\sqrt{x}+2)>0 \Leftrightarrow \sqrt{x}>3 \Leftrightarrow x>9$.

Therefore, $x \in[0 ; 4) \cup(9 ;+\infty)$.