Question
$$n+\frac{n^2}{n}+3-3\frac{n}{n}+3 = 0$$
Answer
$$ \begin{matrix}n_1 = -\dfrac{ 3 }{ 2 }-\dfrac{\sqrt{ 5 }}{ 2 } & n_2 = -\dfrac{ 3 }{ 2 }+\dfrac{\sqrt{ 5 }}{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} n+\frac{n^2}{n}+3-3\frac{n}{n}+3 &= 0&& \text{multiply ALL terms by } \color{blue}{ n }. \\[1 em]nn+n\frac{n^2}{n}+n\cdot3-n\cdot3\frac{n}{n}+n\cdot3 &= n\cdot0&& \text{cancel out the denominators} \\[1 em]n^2+1+3n-3n+3n &= 0&& \text{simplify left side} \\[1 em]n^2+1+3n-3n+3n &= 0&& \\[1 em]n^2+3n+1 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}+3x+1 = 0 $ is a quadratic equation.
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