## Elementary Algebra

{$\frac{-1-\sqrt {13}}{2},\frac{-1+\sqrt {13}}{2}$}
First, we subtract the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$: $n-\frac{3}{n}=-1$ $\frac{n(n)-3(1)}{n}=-1$ $\frac{n^{2}-3}{n}=-1$ $\frac{n^{2}-3}{n}=-\frac{1}{1}$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{n^{2}-3}{n}=-\frac{1}{1}$ $1(n^{2}-3)=-1n$ $n^{2}-3=-1n$ $n^{2}+n-3=0$ Now, we will use the quadratic formula to solve the equation: Step 1: Comparing $n^{2}+n-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=1$, $b=1$ and $c=-3$ Step 2: The quadratic formula is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $n=\frac{-(1) \pm \sqrt {(1)^{2}-4(1)(-3)}}{2(1)}$ Step 4: $n=\frac{-1 \pm \sqrt {1+12}}{2}$ Step 5: $n=\frac{-1 \pm \sqrt {13}}{2}$ Step 6: $n=\frac{-1-\sqrt {13}}{2}$ or $n=\frac{-1+\sqrt {13}}{2}$ Step 7: Therefore, the solution set is {$\frac{-1-\sqrt {13}}{2},\frac{-1+\sqrt {13}}{2}$}.