Odpowiedź :
[tex]a_{n} = n^{2}-2n-24\\\\n^{2}-2n-24 < 0\\\\a = 1, \ b = -2, \ c = -24\\\\\Delta = b^{2}-4ac = (-2)^{2}-4\cdot1\cdot(-24)=4+96 = 100\\\\\sqrt{\Delta} = \sqrt{100} = 10\\\\n > 1\\\\n_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-2)-10}{2\cdot1} = \frac{2-10}{2}=\frac{-8}{2} = -4 \ \ \notin D\\\\n_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-2)+10}{2} = \frac{2+10}{2} = \frac{12}{2} = 6[/tex]
a > 0, to ramion pataboli skierowane do góry
[tex]n \in (1; 6)\\\\\boxed{n = 5}[/tex]
[tex]\underline{Odp. \ B.}[/tex]