[tex]n \in C \\n > 1\\\\\sqrt{2n} < n < \sqrt{5n}\\2n < n^2 < 5n\\\\n^2 > 2n\\n^2-2n > 0\\\\\Delta=(-2)^2-4*1*0=4\\\sqrt{\Delta}=2\\n_1=\frac{2-2}2=0\\n_2=\frac{2+2}2=\frac42=2\\n\in (-\infty; 0)U(2; \infty)\\(-\infty; 0) \notin D\\\\n\in (2; \infty)\\\\n^2 < 5n\\n^2-5n < 0\\\Delta=(-5)^2-4*1*0=25\\\sqrt{\Delta}=5\\n_3=\frac{5-5}2=0\\n_4=\frac{5+5}2=5\\\\n\in(0; 5)\\(0; 1) \notin D\\n\in (1; 5)\\\\n\in (2; 5)\\\bold{\underline{n = \{3, 4\}}}[/tex]
