Answer
$x=-\frac{1}{3},x=-\frac{1}{5}$
Work Step by Step
Let's find the solutions as follows.
$\begin{align}\frac{1}{x^2}+\frac{8}{x}+15&=0\\\frac{1+8x+15x^2}{x^2}&=0\\15x^2+8x+1&=0\\15x^2+3x+5x+1&=0\\3x(5x+1)+1(5x+1)&=0\\(5x+1)(3x+1)&=0\\x&=-\frac{1}{5} \text{ or } x=-\frac{1}{3}\end{align}$
Now to check substitute $x=-\frac{1}{5}$ and $x=-\frac{1}{3}$ in original equation.
$\frac{1}{\left(\frac{-1}{5}\right)^2}+\frac{8}{\frac{-1}{5}}+15=25-40+15=0$
$\frac{1}{\left(\frac{-1}{3}\right)^2}+\frac{8}{\frac{-1}{3}}+15=9-24+15=0$
Hence the solutions are $x=-\frac{1}{3},x=-\frac{1}{5}$.
