the ODE is:
$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$
with $$x=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$
you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$
$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$
hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$
you can obtain the time where the velocity is maximum by solving this equation
$$\frac{d}{dt} v(t)=0$$
for t, $~\Rightarrow~t_m=t$ and
$~|v_m|=\frac{\omega\,x_0}{e}$