4th kinematic equations for constant acceleration Ive been given the 4 kinematic equations for constant acceleration. The fourth being:
$$s=ut+\frac{1}{2}at^2.$$
If rearranged it forms the quadratic equation
$$at^2+2ut-2s=0.$$
But that means that $t$ has 2 values. 


*

*Will one of them always be negative? 
So only one value is realistically possible?

*And how could you rearrange it to get $t$ on it's own? 
Would that be using the quadratic formula?
 A: The equation describes parabolic motion, if $a\neq 0$ is a non-zero constant acceleration, which I will assume from now on. If you think about it, your solution provides an answer to the question: at what time does the object is in the position $s$? [A note on notation: Traditionally, the letter $s$ denotes distance (I guess from the German word "Strecke"), which by definition is a non-negative quantity, but your formula makes more sense, if we interpret $s$ as a position $x$, which can also be negative.]
$$\frac{1}{2}at^2+ut-x=0$$
$$t_1 = \frac{-u-\sqrt{D}}{a}$$
$$t_2 = \frac{-u+\sqrt{D}}{a}$$
where $D := u^2+2ax$. Let's think about it for a moment, and see what answers we get by varying $x$.
Case $D<0$:
The discriminant is negative, there are no solutions, therefore at no time your object will have that position.
Case $D=0 \Leftrightarrow x=-\frac{u^2}{2a}$:
The discriminant is zero, there is only one solution which is the "top" ("bottom") point reached by the object,if $a<0$ ($a>0$), respectively.
Case $D>0$:
The discriminant is positive and there are two solutions. This means that the object will reach that position twice, once going "up" and once going "down" the parabola.
Now, will one of them always be negative? Not necessarily.
Case $ax > 0$: One positive and one negative.
Case $ax < 0$ and $\frac{u}{a}>0$: Two negative.
Case $ax < 0$ and $\frac{u}{a}<0$: Two positive.
Case $x = 0$ and $\frac{u}{a}>0$: One zero and one negative.
Case $x = 0$ and $\frac{u}{a}<0$: One zero and one positive.
A: You are correct about how to solve for t, it's a quadratic equation of the form $at^2 + bt + c = 0$, so the solution is
$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
As to your first question, solving for t means you are looking for the exact time at which the position of the object is s. The velocity at time $t=0$ is u, but the acceleration a is constant, so that means that before time $t=0$, the velocity was less than u and was accelerating up to u. If you look far enough back in the past, the velocity must have been negative, and so the object would have passed through the point s sometime. Therefore, both solutions are valid.
However, when we pose a problem like this, it is usually implicit that we are only interested in what happens starting from $t=0$, unless we specifically mention otherwise. So you can usually ignore the negative solution.
A: I we take the initial velocity as zero ($u=0$) then we have $s=1/2at^2$ from which a trivial solution provides two displacement $s_1$ and $s_2$ depending on if we take the positive or negative root.
Drawing a graph of $s - v t^2$ gives a parabola and our two solutions appear somewhere on this curve.  This is where we need to disentangle the mathematical explanation from the physical explanation.  Mathematically they are both valid and it appears our equation is very useful but physically we can choose to prefer one solution to the other.
In this case I would take the positive root as this describes the position at some positive time, where the negative root gives the position at some negative time.  Taking time to be positive and going forward isn't always correct but it can help differentiate between mathematically equivalent answers.
A: When you throw a ball into the air, there are two times it will have a certain height $s$. Once going up and once coming down. That is what the two times represent. 
Both are valid.
