How can I find these velocities without using the quadratic formula? 
If a ball is thrown vertically upward with a velocity of $160 \text{ ft/s}$, then its height after t seconds is $s = 160t − 16t^2$.
a) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way up?
  b) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way down?

$$\begin{align*}
384 &= 160t - 16t^2\\ 
16t^2 - 160t + 384 &= 0\\  
16 (t^2 - 10t + 24) &= 0
\end{align*}$$ 
I can't factor the above... and I'm not supposed to used the quadratic formula. Am I stupid or is this unsolvable without without the formula?
 A: $$16(t^2 - 10 t + 24) = 0 \iff 16(t-4)(t-6) = 0 \implies t = 4,\;\; t = 6$$
Heading up: $384$ feet at time $t = 4$, 
Descendng down after reaching maximum height: $384$ ft. at time $t = 6$.
Note: $$(-4)\cdot (-6) = + 24;\quad -4 + - 6 = -10$$
Noticing those facts allow you to deduce that the factors must be $$\;t^2 - 10 t + 24 = (t - 4)(t - 6)$$
A: $$\text{So, }v=\frac{ds}{dt}=160-32t$$
$$\text{Now, }160t-16t^2=384\implies t^2-10t+24=0\implies t=4,6$$
The smaller value of $t$ will occur during upward movement
A: If we know that the formula for motion under uniform acceleration is
$s = v_0 t + \frac12 at^2,$ where $v_0$ is the initial velocity and $a$ is the acceleration, then a glance at the formula $s = 160 t - 16t^2$
tells us that $v_0=160$ (which was given in the problem statement anyway)
and $a = -32.$
It therefore takes $(-160)/(-32) = 5$ seconds to reduce the upward velocity from $160$ to $0.$ 
At the end of $5$ seconds, the ball's height is $160(5)-16(5^2) = 400$ feet
and it is at the highest point it will reach.
It follows that the plot of the ball's height over time is a parabola
with vertex at $t=5$ and formula $s = 400 - 16(t-5)^2.$
We plug in $s = 384,$ that is, $384 = 400 - 16(t-5)^2.$
Subtracting $400,$ we have $-16 = - 16(t-5)^2,$
so we just need to divide by $-16$ and then take square roots to determine
that $t - 5 = \pm 1.$
So its velocity on the way down is just the velocity an object has after falling $1$ second from a rest position (which is a negative value), and to get the velocity on the way up we just reverse the sign.
Or we can do it "the hard way" and compute the velocities at 
$t=5-1$ and $t = 5 + 1.$

If we know about potential and kinetic energy, however, then it's just plug-and-chug.
The kinetic energy at ground level is 
$\frac12mv_0^2 = \frac12m(160^2) = 12800m,$ 
where $m$ is the mass of the ball.
Setting potential at the ground level to zero, the total energy is
therefore $12800m.$
The potential energy at $384$ feet is $mgs = m\times32\times384 = 12288m.$
Since kinetic plus potential is constant, that means
$12288m + KE_{384} = 12800m,$
where $KE_{384}$ is the kinetic energy at $384$ feet.
So $KE_{384} = 12800m - 12288m = 512m = \frac12mv_{384}^2,$
where $v_{384}$ is the velocity at $384$ feet;
so $v_{384}^2 = 1024.$ Taking square roots, $v_{384} = \pm 32.$
Lots of things get simpler when we know how to apply conservation of energy.
