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, $~\Rightarrow~t_m=t$