The motion of the car is described by the kinematic equation for displacement under constant acceleration:

$$\Delta x = v_0 t + \frac{1}{2}at^2$$

where $\Delta x$ is the displacement, $v_0$ is the initial velocity, $a$ is the constant acceleration, and $t$ is the time.

To solve for the time $t$, we rearrange this equation into the standard quadratic form $At^2 + Bt + C = 0$:

$$\left(\frac{1}{2}a\right)t^2 + (v_0)t - \Delta x = 0$$

Applying the quadratic formula, $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, gives the general solution for time:

$$t = \frac{-v_0 \pm \sqrt{v_0^2 - 4\left(\frac{1}{2}a\right)(-\Delta x)}}{2\left(\frac{1}{2}a\right)}$$ $$t = \frac{-v_0 \pm \sqrt{v_0^2 + 2a\Delta x}}{a}$$

Now, we substitute the given values: $v_0 = 5$ m/s, $a = -0.4$ m/s², and $\Delta x = 30$ m.

$$t = \frac{-(5) \pm \sqrt{(5)^2 + 2(-0.4)(30)}}{-0.4}$$ $$t = \frac{-5 \pm \sqrt{25 - 24}}{-0.4}$$ $$t = \frac{-5 \pm \sqrt{1}}{-0.4}$$ $$t = \frac{-5 \pm 1}{-0.4}$$

This yields two positive, physically valid solutions for the time:

$$t_1 = \frac{-5 + 1}{-0.4} = \frac{-4}{-0.4} = 10 \text{ s}$$ $$t_2 = \frac{-5 - 1}{-0.4} = \frac{-6}{-0.4} = 15 \text{ s}$$

The first time, $t_1$, corresponds to the car reaching 30 m while moving forward. The second time, $t_2$, corresponds to the car reaching its maximum displacement, reversing, and passing 30 m again on its way back.