## Elementary Algebra

{$\frac{3-\sqrt {137}}{8},\frac{3+\sqrt {137}}{8}$}
First, we subtract the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$: $n-\frac{2}{n}=\frac{3}{4}$ $\frac{n(n)-2(1)}{n}=\frac{3}{4}$ $\frac{n^{2}-2}{n}=\frac{3}{4}$ Now, we cross multiply the two fractions in order to create a linear equation: $\frac{n^{2}-2}{n}=\frac{3}{4}$ $4(n^{2}-2)=3n$ $4n^{2}-8=3n$ $4n^{2}-3n-8=0$ Now, we will use the quadratic formula to solve the equation: Step 1: Comparing $4n^{2}-3n-8=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=4$, $b=-3$ and $c=-8$ 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: $n=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(4)(-8)}}{2(4)}$ Step 4: $n=\frac{3 \pm \sqrt {9+128}}{8}$ Step 5: $n=\frac{3 \pm \sqrt {137}}{8}$ Step 6: $n=\frac{3-\sqrt {137}}{8}$ or $n=\frac{3+\sqrt {137}}{8}$ Step 7: Therefore, the solution set is {$\frac{3-\sqrt {137}}{8},\frac{3+\sqrt {137}}{8}$}.