Solving for the interval of time during which the height of a thrown ball is at least "h" feet In dealing with inequalities I've run into a certain peculiarity which I am currently unable to explain.
The example: Find the interval of time during which the ball is at least 32 feet above ground.
h = -16t^2 + 16t + 128 // Height of the ball in feet.

-16t^2 + 16t + 128 >= 32
-16t^2 + 16t + 96 >= 0
-16(t^2 - t - 6) >= 0
-16(t+2)(t-3) >= 0 // At this point everything is going as planned.


// Now I have a choice to make
(-16t - 32)(t - 3) >= 0 // This does not work  OR
(t + 2)(-16t + 48) >= 0 // This does work


// If I choose option 1, the relational operators are incorrect.
-16t - 32 >= 0      and      t - 3 >= 0
-16t >= 32                   t >= 3
t <= -2


// If I choose option 2, the relational operators are correct.
t + 2 >= 0          and      -16t + 48 >= 0
t >= -2                      -16t >= -48
                             t <= 3

Now, when looking at a graph it becomes obvious that the ball is at least 32 feet above ground during the interval [0, 3] (assuming time is not negative). Therefore, option 2 provides the correct relations for t, while option 1 inverts the relations.
What I don't understand is why this is happening, since multiplication is an associative operator. Seems to me that it shouldn't matter whether the -16 is multiplied into the first factor or the second, and yet it does. I would love to know why, so that I might circumvent this issue the next time around.
 A: It comes from your line where you require both 
$$\begin{align}
-16t - 32 &> 0 \\
t - 3 &> 0
\end{align} $$
As you saw, this is not sensible. But you might instead require both
$$\begin{align}
-16t - 32 &< 0 \\
t - 3 &< 0
\end{align} $$
And this does lead to the correct answer. Recall that the final answer is when either of these two cases are true.
A: Absorbing the $-16$ into one of the terms was unnecessary, as we will see  later.  But let's start from your actual calculation.
Look at the version that "doesn't work", namely
$$(-16t-32)(t-3) \ge 0.$$
I have rewritten $\lt$ as $\le$, and $\gt$ as $\ge$, because "at least $32$ feet" means $32$ or more.
The displayed inequality is true if (i) $-16t-32$ and $t-3$ are both $\ge 0$ OR (ii) $-16t-32$ and $t-3$ are both $\le 0$.
Note that $-16t-32 \ge 0$ iff $-16t \ge 32$ iff $t\le -2$.  
The condition $t-3\ge 0$ can be rewritten as $t\ge 3$.  So our analysis of case (i) shows that it holds iff $t\le-2$ and $t \ge 3$. But these are clearly incompatible, so case (i) cannot hold.
Or else, as you observed, it is implicit in the problem that $t \ge 0$, so $t \le -2$ is physically irrelevant. 
For case (ii), look first at $-16t-32\le 0$. Rewrite this as $-16t \le 32$, and then $t\ge -2$. Rewrite the condition $t-3\le 0$ as $t \le 3$.  So as far as the formula is concerned, everything is OK if $-2\le t\le 3$.  But $t \ge 0$, so we conclude that the answer is $0 \le t \le 3$.
Thus we got a complete and correct analysis out of the "doesn't work."  
However, let's start again from $-16(t+2)(t-3) \ge 0$.  
This is equivalent to $(t+2)(t-3) \le 0$.
That is true if (i) $t +2 \le 0$ and $t-3\ge 0$ OR (ii) $t+2\ge 0$ and $t-3 \le 0$. (A product is $\le 0$ if one term is $\le 0$ and the other is $\ge 0$.)
Case (i) is physically irrelevant.  Anyway, it yields the incompatible $t \le -2$ and $t \ge 3$.
Case (ii) yields $t \ge -2$ and $t \le 3$.  For physical reasons this should be corrected to $0 \le t \le 3$.   
A: from your  inequality   -16*t^2+16*t+128>32  there is one question if at least  32 does it  mean that not only >32 but also >=32? so rewrite your equation like this
-16*t^2+16*t+128>=32

-16*t^2+16*t+96>=0   now  dive it by -16 and change > by < so  we will have
t^2-t-6<=0  solution is  t1=3 t2=-2 ,but because feet can't be negative answer will be  [0 3]
