A concave mirror has focal length $f=-7.5\,\text{cm}$. An object is placed at $u=-5\,\text{cm}$. The image formed is

Using the mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, we get $\frac{1}{v}-\frac{1}{5}=-\frac{1}{7.5}$, so $v=+15\,\text{cm}$. A positive image distance for this case means the image is behind the mirror, hence virtual. Magnification $m=-v/u=-15/(-5)=+3$, so the image is erect and magnified.