Question 322602
For part A, just set the given quadratic equal to 200 and solve for t.

In other words, solve {{{-4.9t^2+86t-197.9=0}}}

You will get two solutions. One for when the rocket is going up and reaches 200 m, and the other when it is on its way down and descends to 200 m. I assume they mean the former (when it is ascending).

For part B, set the given quadratic equal to 0 and solve for t.

{{{-4.9t^2+86t+2.1=0}}}

As an added bonus, find the maximum height of the rocket and how long it takes to get there.

You can differentiate the given quadratic or use t=-b/(2a) 

-86/(2(-4.9))=8.775 seconds

The max height is then found by plugging back into the quadratic:

-4.9(8.775)^2+86(8.775)+2.1=379.45 m

Note that if we double 8.775, we get the same solution as in part B because the max height is halfway to the point where it lands. Very close, due to rounding.