Answer
The solutions are $x=-\frac{1}{2}$ and $x=-\frac{1}{4}$
Work Step by Step
$\bigg(\dfrac{x+1}{x}\bigg)^2+4\bigg(\dfrac{x+1}{x}\bigg)+3=0\hspace{0.7cm} \color{blue}{\text{Given equation}}$
$W^2+4W+3=0\hspace{3.3cm} \color{blue}{\text{Let } W=\dfrac{x+1}{x}}$
$(W+1)+(W+3)=0\hspace{2.3cm} \color{blue}{\text{Factor}}$
$W+1=0\hspace{0.5cm}or\hspace{0.5cm}W+3=0\hspace{1.3cm} \color{blue}{\text{Zero-Product Property}}$
$W=-1\hspace{1.5cm}W=-3\hspace{1.99cm} \color{blue}{\text{Solve}}$
Now change $W$ back into the correspondimg values of $x$
$\dfrac{x+1}{x}=-1\hspace{1.5cm}\dfrac{x+1}{x}=-3$
$x+1=-x\hspace{1.5cm}x+1=-3x$
$-2x=1\hspace{1.5cm}4x+1=-1$
$x=-\frac{1}{2}\hspace{1.5cm}x=-\frac{1}{4}$
The solutions are $x=-\frac{1}{2}$ and $x=-\frac{1}{4}$