Two different time periods for a movement with constant acceleration? I'm studying for my physics exam and I keep running into the same problem. It's so specific I have no idea how to phrase it in a Google or stack exchange search, and I've already wasted 2 hours on it. So apologies if this is too simple a question.
The problem I will use as an example is complicated but I will simplify it (a lot) to address the confusion I'm having:

How long will it take for an object moving at $5.0~\mathrm{m/s}$ and slowing down by $0.5 ~\mathrm{m/s^2}$ to travel a distance of $17$ meters?

With the formula:
$$x_\mathrm{final} = x_\mathrm{initial} + v_\mathrm{start}*dt + a / 2*dt^2$$
Assuming $x_\mathrm{initial} = 0$ and $x_\mathrm{final} = 17 ~\mathrm{m}$, I can use the $dt$ or "change in time" as an unknown, and can call it $t$.
This gives the function:
$$-0.25t^2 + 5t - 17 = 0$$
When solving this function I get:
$$D = 5^2 - 4*(-0.25)*(-17) = 8\quad ,\\
t1 = \frac{-5 + 8^{1/2}}{-0.5}\quad,\\
t2 = \frac{-5 - 8^{1/2}}{-0.5}$$
The two answers rounded off are: $t1 = 4.3431~\mathrm{seconds}$ and $t2 = 15.6569 ~\mathrm{seconds}$.
How the is it possible that an item with a constant acceleration can displace over the same distance in two different time periods? This is really confusing to me. I would expect one of the answers to be a negative number, which usually is the case in these types of situations. Then it is logical that negative time is not possible and that the positive number is the only answer. Why is that no longer the case here?
At first I thought I had used the wrong $+$ and $-$ symbols for my given values but I made like 20 different drawings and made the calculation over and over again and I don't understand what I am doing wrong. If we put the movement on the $x$-axis, the movement is going from $x=0$ to $x =17$. The speed must be positive so: $v=5.0$. The acceleration is slowing the movement down so: $a = -0.5$.
When I plug this in to the formula that I have triple checked from the book lying right next to me I still get the same answer.
Again, I apologize for asking such a stupid question because this is just elementary physics and I have never had these issues in high school. I don't understand why this is suddenly so confusing for me. I study computer science now and this is one of our minor subjects, so I don't put a lot of thought into the subject throughout the year.
I'm guessing that I am somehow using the formula in a wrong way because if I solve the function: $-0.25t^2 + 5t + 17 = 0$ then I do find a correct answer but I don't get why i should be using $17$ as a positive number instead of a negative one.
 A: That you get two answers for time $t$ is perfectly fine!

How the is it possible that an item with a constant acceleration can displace over the same distance in two different time periods?

Remember, that the $17$ meters you put in is not distance, but a position. The object passes this position at $t1$, then it decelerates, speeds down, gets a negative speed, comes back, and passes that position again at $t2$.
See my plot here:

The blue line is the 17 meters. Position $x$ at the y-axis. There are two perfectly fine cases where your equation is true: at the intersections between the two lines.
A: There's nothing wrong here about the equation giving you two instants. Say, your object is moving in the positive $x$ direction, it started with $v=5$ , but it's slowing down. "Slowing down" basically means that the direction of the acceleration is opposite to your velocity, that is, $a$ is in the negative $x$ direction. The object will cover a certain amount of distance before it stops. Now at $t=4.3$, this object would have already covered $17$ meters, then it'll continue covering some additional distance before it stops.
Then what happens after it stops? that is, after $v=0$?, well, it will start to speed up and move in the negative $x$ direction($a$ is negative, remember?). So that at $t=15.6$, it will cover another $17$ meters.
A: I haven't checked your calculations, but the reason you can have 2 solutions is as follows.  After 4.3 seconds, the object has travelled 17m from its start point, say heading along positive x axis.  After 10 seconds it reaches a halt and starts to go backwards, and after 15.6 seconds, it's back at  the point 17m from the start.
