= Solution
For $V(\phi)=\lambda\phi^4$, the <slow-roll approximation> gives
$$
H^2\simeq\frac{8\pi G}{3}\lambda\phi^4,
\qquad
3H\dot\phi\simeq-4\lambda\phi^3.
$$
Taking the expanding branch $H>0$ and using $H=\sqrt{8\pi G\lambda/3}\,\phi^2$ gives the <first-order linear differential equation>
$$
\dot\phi=-\sqrt{\frac{2\lambda}{3\pi G}}\,\phi.
$$
Thus the <quartic-potential slow-roll solution> with the normalization used in this question is
$$
\boxed{
\phi(t)=\phi_0
\exp\left[-\sqrt{\frac{2\lambda}{3\pi G}}(t-t_0)\right]}.
$$
If $\phi_0<0$, the field remains negative and $\dot\phi>0$, so it rolls toward the potential minimum at zero.
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