Elementary Algebra

{$\frac{3+\sqrt {33}}{4},\frac{3-\sqrt {33}}{4}$}
First, we write $2n=3+\frac{3}{n}$ as $3+\frac{3}{n}=2n$. Then, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$: $3+\frac{3}{n}=2n$ $\frac{3(n)+3(1)}{n}=2n$ $\frac{3n+3}{n}=2n$ $\frac{3n+3}{n}=\frac{2n}{1}$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{3n+3}{n}=\frac{2n}{1}$ $2n(n)=1(3n+3)$ $2n^{2}=3n+3$ $2n^{2}-3n-3=0$ Now, we will use the quadratic formula to solve the equation: Step 1: Comparing $2n^{2}-3n-3=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find: $a=2$, $b=-3$ 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: $n=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(2)(-3)}}{2(2)}$ Step 4: $n=\frac{3 \pm \sqrt {9+24}}{4}$ Step 5: $n=\frac{3 \pm \sqrt {33}}{4}$ Step 6: $n=\frac{3+\sqrt {33}}{4}$ or $n=\frac{3-\sqrt {33}}{4}$ Step 7: Therefore, the solution set is {$\frac{3+\sqrt {33}}{4},\frac{3-\sqrt {33}}{4}$}.