What will be the nature of the image formed when an object is placed at $20\mathit{ cm}$ in front of a concave mirror of a focal length of $12\mathit{ cm}$?

The nature of the image formed will be magnified, real and inverted when an object is placed at $20\mathit{ cm}$ in front of a concave mirror of focal length $12\mathit{ cm}$.
Given,
$u= -20\mathit{ cm}$
$f= -12\mathit{ cm}$    (Negative: as the mirror is concave)
By using the mirror formula, we can calculate the position of the image by substituting the values of $u$ and $f$ which are $-20$ and $-12$ respectively. 
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{1}{(-12)}=\frac{1}{v}+\frac{1}{\left(-20\right)}$
$\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{12}$
$\Rightarrow \frac{1}{v}=\frac{3-5}{60}$
$\Rightarrow \frac{1}{v}=\frac{-1}{30}$
$\Rightarrow \mathit{ v}=-30\mathit{ cm}$
Now, substituting the values of $u$ and $v$ in magnification formula we get,
$m=-\frac{v}{u}$
$\Rightarrow \mathit{ m}=-\frac{(-30)}{(-20)}$
$\Rightarrow \mathit{ m}=-1.5$ $-\mathit{ve}$ sign indicates that the image formed will be real and inverted.
So, the image formed will be real, inverted and magnified.