Finding out minimum values of $\phi$

Finding out minimum values of $\phi$

The Lagrangian for $\phi^4$ theory, $$\mathcal{L}=
\frac{1}{2}(\partial_\mu \phi)^2 - \frac{1}{2} m^2\phi^2 -
\frac{\lambda}{4}\phi^4 ,$$
The Kinetic energy K \begin{equation*} T =\frac{1}{2}(\partial_\mu
\phi)^2, \end{equation*} and the potential energy $ V(\phi)$ is,
\begin{equation*} V(\phi) = \frac{1}{2} m^2\phi^2 +
\frac{\lambda}{4}\phi^4. \end{equation*} What my intention is to get the
minimum value during symmetry breaking. To find out minimum we need to
take derivative. Taking derivative with respect to $\phi$, we get,
\begin{align*} \frac{\partial V }{\partial \phi} &= m^2 \phi+ \lambda
\phi^3 \\ &=\phi (m^2 + \lambda \phi^2). \end{align*} Can someone explain
the calculus for determining the unstable minimum?
EDIT
I mean why $\frac{\partial^2 V }{\partial \phi^2} =0$ to show that $m^2>0$?