Maybe you started by asking for something like this...
ASSUMPTION. We start at time $t = 0$ and consider values $t \ge 0.$
We end when the rocket is on the ground, that means height $0,$ so we must have $h(0) = 0$. We need to find $t$ such that $h(t) = 0.$
GIVEN. $h(t) = 160 - 16t^2$. The rocket must start at its highest
point because $h(0) = 160$ and the height $h(t)$ only decreases from there on for larger values of $t$.
FORMULA. From some Comments, I think you want to use the quadratic formula to solve
the equation to find $t$ such that $h(t) = 0.$
The equation starts as $0 = 160 - 16t^2.$ To use the quadratic
formula it needs to be in the form $at^2 + bt + c = 0.$
So that's $-16t^2 + 0t + 160 = 0,$ where $a = -16,\,b=0,\,$and
$c = 160.$
SOLUTION: Then the quadratic formula becomes
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
= \frac{0 \pm \sqrt{0 - 4(-16)(1600)}}{2(-16)}
= \frac{\pm \sqrt{10240}}{-32} = \mp 3.162278.$$
Of the two values $t = -3.1623$ and $t = 3.1623,$ only
the positive one is useful, because we've assumed from
the start that $t \ge 0.$ So the rocket crashes to
the ground after 3.1623 seconds. This is the same as $\sqrt{10},$
as mentioned in the Comments.
Note: I can understand how someone who considers himself
'not very good at math' would want the security of using
the quadratic formula--which always works--even if it turns out to have put you
through a little messy arithmetic in this case. I hope
you take a good look at what @Kenneth was saying in his
Comments. That is a simpler way to solve this particular
quadratic equation (in which $b = 0$). Technically,
the answer there is really $\pm\sqrt{10},$ but the $\pm$
wasn't mentioned in the Comments because everyone
was assuming that $t \ge 0$ from the start.
If I remember my history of math correctly it must
have been a bit over 300 years ago that Isaac Newton
was first playing around with this and related equations.
I'll bet he would have been willing to pay a huge price
for your hand calculator.
As discussed in the Comments, the other part is to
give the height at $t = 2$ sec. That's just
$t(2) = 160 - 16(2^2) = 160 - 64 = 96$ feet high.
I don't know how good you are at graphing and interpreting
graphs, but I've put one below of the curve $h(t) = 160 - 16t^2$
for $t$ between $0$ and $\sqrt{10} = 3.1623.$ Orange lines
show that the height is $96$ feet at time $2$ seconds.
You've ignored the multiplication of initial velocity by time of flight. The total distance above ground is given by $$3+14t-5t^2$$which then must be solved for equality to zero distance above ground, and account for common sense.
Gravitational acceleration creates a change in distance of $$-5t^2$$ if we use 1 significant figure. The only thing affecting the gravitational changes is gravity itself, which has no dependence on velocity (at least in his scale).
The initial velocity changes distance by a separate factor,$$v_it$$
which does not depend on gravity, hence these terms are separate in the equation. The constant term is straightforwardly the height at time zero.
Best Answer
For the second question,
The ball is moving downward from a certain height to ground zero.
So, the initial velocity of 40 feet/sec will be the final velocity in the downward direction.
Because, when you throw a ball upward vertically with an initial velocity of u, at the highest point, the initial velocity becomes zero (or, momentarily rest) and then, it travels downward with final velocity v, but having the same magnitude as u.
So, using the following equations, one can find the solution to the question (b).
Step 1) Find the time taken to reach the maximum height that the ball will reach vertically upward.
Step 2) Once it has reached that height, then find the time taken to reach the ground.
So, we know, v = u + at. In this context, a = g (gravitational acceleration).
So, for Step 1) t1 = u sin(theta) / g = u/g [ v = u Sin(theta) - gt1 ; since v = 0, therefore, t1 = u sin(theta) / g ] [Here, theta = 90, since the ball is thrown vertically upward at an angle of 90 degree with the horizontal].
And for Step 2) t2 = u sin(theta) / g = u/g [ v = u Sin(theta) + gt2 ; since v = u sin(theta) and u = 0, therefore, t2 = u sin(theta) / g ] [Here, theta = 90, since the ball is thrown vertically upward at an angle of 90 degree with the horizontal].
Therefore, the Total time of flight for the ball after being hit is as follows.
t = t1 + t2 = (2*u)/g [ g = 32.1741 feet per second squared]
= (2*40)/(32.1741)
= 2.4865 seconds.