+128408 I assume we have....
1 / [2n^2] + 5 / [2n] = n - 2 /[n^2] multiply through by 2n^2....this gives...
1 + 5n = 2n^3 - 4 rearrange
2n^3 - 5n - 5 = 0 this isn't factorable.....the only "real" solution occurs at about n = 1.946
See the graph here......https://www.desmos.com/calculator/q449wlwqlg
$${\mathtt{2}}\left[{{n}}^{{\mathtt{3}}}\right]{\mathtt{\,-\,}}{\mathtt{5}}{n}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,-\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{1.945\: \!510\: \!206\: \!541\: \!832\: \!2}}\\
\end{array} \right\}$$
+128408 I assume we have....
1 / [2n^2] + 5 / [2n] = n - 2 /[n^2] multiply through by 2n^2....this gives...
1 + 5n = 2n^3 - 4 rearrange
2n^3 - 5n - 5 = 0 this isn't factorable.....the only "real" solution occurs at about n = 1.946
See the graph here......https://www.desmos.com/calculator/q449wlwqlg
$${\mathtt{2}}\left[{{n}}^{{\mathtt{3}}}\right]{\mathtt{\,-\,}}{\mathtt{5}}{n}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{n}} = {\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{17}}}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{4}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,-\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{\,-\,}}{\mathtt{0.972\: \!755\: \!103\: \!270\: \!916\: \!1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.582\: \!028\: \!756\: \!006\: \!450\: \!1}}{i}\\
{\mathtt{n}} = {\mathtt{1.945\: \!510\: \!206\: \!541\: \!832\: \!2}}\\
\end{array} \right\}$$
