Given: $\mathit{u}=12\mathrm{cm}$ (negative), $\mathit{f}=8\mathrm{cm}$ (positive)
We know, by lens formulae,
$\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}$
$\frac{1}{\mathit{v}}=\frac{1}{\mathit{u}}+\frac{1}{\mathit{f}}$
or $\frac{1}{\mathit{v}}=\frac{1}{\mathit{u}}+\frac{1}{\mathit{f}}$
or $\frac{1}{\mathit{v}}=\frac{1}{-12}+\frac{1}{8}$
$\frac{1}{\mathit{v}}=\frac{-2+3}{24}=\frac{1}{24}$

Thus, the image is formed at a distance of 24 $\mathrm{cm}$ behind the lens.
We know, $\mathit{m}=\frac{\mathit{v}}{\mathit{u}}=\frac{24}{-12}=-2$
Thus, the image is Real, inverted and magnified in size.