I don't understand exactly what you mean, so there are two options
1.
[tex]\displaystyle \frac{r+3}{r} =-4\\\\r+3=-4r\\\\3=-5r\\\\r=- \frac{3}{5} [/tex]
2.
[tex]\displaystyle r+\frac{3}{r} =-4\\\\r^2 +3=-4r\\\\r^2+4r+3=0\\\\r_{1,2}= \frac{-4\pm \sqrt{16-12} }{2}= \frac{-4\pm 2}{2} =(-3), (-1)[/tex]
