Answer
{$-4,1$}
Work Step by Step
First, we subtract the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$:
$n-\frac{4}{n}=-3$
$\frac{n(n)-4(1)}{n}=-3$
$\frac{n^{2}-4}{n}=-3$
$\frac{n^{2}-4}{n}=-\frac{3}{1}$
Now, we cross multiply the two fractions in order to create a quadratic equation:
$\frac{n^{2}-4}{n}=-\frac{3}{1}$
$1(n^{2}-4)=-3n$
$n^{2}-4=-3n$
$n^{2}+3n-4=0$
Now, we use rules of factoring trinomials to solve the equation:
$n^{2}+3n-4=0$
$n^{2}-1n+4n-4=0$
$n(n-1)+4(n-1)=0$
$(n-1)(n+4)=0$
$(n-1)=0$ or $(n+4)=0$
$n=1$ or $n=-4$
Therefore, the solution is {$-4,1$}.
