$$\frac{1}{n+3}+\frac{5}{n^2-9} = 0$$
Answer
$$ \begin{matrix}n_1 = -2 & n_2 = -3 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{n+3}+\frac{5}{n^2-9} &= 0&& \text{multiply ALL terms by } \color{blue}{ (n+3)(n^2-9) }. \\[1 em](n+3)(n^2-9)\cdot\frac{1}{n+3}+(n+3)(n^2-9)\cdot\frac{5}{n^2-9} &= (n+3)(n^2-9)\cdot0&& \text{cancel out the denominators} \\[1 em]n^2-9+5n+15 &= 0&& \text{simplify left side} \\[1 em]n^2+5n+6 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}+5x+6 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
This page was created using
Equations Solver