How can I find these velocities without using the quadratic formula?

Question

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?

Answer 1

$$16(t^2 - 10 t + 24) = 0 \iff 16(t-4)(t-6) = 0 \implies t = 4,\;\; t = 6$$

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 $$\;t^2 - 10 t + 24 = (t - 4)(t - 6)$$

Answer 2

$$\text{So, }v=\frac{ds}{dt}=160-32t$$

$$\text{Now, }160t-16t^2=384\implies t^2-10t+24=0\implies t=4,6$$

The smaller value of $t$ will occur during upward movement