Can anyone answer this with the steps of Assumption > Given > Formula > Solution? The height in feet of a bottle rocket is given by the function $h(t)=160-16t^2$, where $t$ is the time in seconds. How long will it take for the rocket to return to the ground? What is the height of the rocket after 2 seconds?
 A: 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.

