Question 974255
If you are not given the function that represents height as a function of time, this is a physics problem, rather than an algebra problem.
A friendly physics problem would say that you can use {{{g=10}}}{{{m/s^2}}} .
Lacking such show of friendship, we are obligated to use the cumbersome {{{g=9.8}}}{{{m/s^2}}} that is a better approximation of {{{g}}} on/near Earth's surface.
 
If we could use {{{g=10}}}{{{m/s^2}}} ,
parts a) and b)are easy mental math:
a) The rocket reach its maximum height when its upwards velocity becomes {{{0}}} ,
and if it starts at {{{"25 m / s"}}} ,
decreasing at a rate of {{{"10 m /"}}}{{{s^2="10 m / s /s"}}} ,
that would take {{{"25 m / s"/"10 m / s /s"=2.5s}}}
b)Since the upwards velocity was decreasing linearly (at a constant rate),
the average upwards velocity for the upwards flight is the average of the initial and final velocities,
{{{("25 m / s"+"0 m / s")/2="12.5 m / s"}}} .
during the {{{2.5s}}} upwards flight with an average velocity of {{{"12.5 m / s"}}} ,
the rocket traveled {{{2.5s("12.5 m / s")=31.25m}}} .
Since the rocket started its upwards flight  at a height of {{{10m}}} ,
its maximum height is {{{10m+31.25m=41.25m}}} .
 
If you must use {{{g=9.8}}}{{{m/s^2}}} ,
the calculations are the same,
but I suggest using a calculator.
 
For the rest,
{{{h}}}= height of the rocket above the ground, in m
{{{h[0]}}}= initial height of the rocket above the ground, in m{{{t}}}= time since launch, in seconds
{{{v[0]}}}= initial velocity, in m/s
{{{g}}}= acceleration of gravity, in m/s/s.
{{{v}}}= upwards velocity, in m/s
{{{v=v[0]-gt}}}
The average velocity in the interval [0,t] is
{{{(v[0]+v)/2=(v[0]+v[0]-gt)/2=v[0]-gt/2}}}
and consequently, the gain in height in the interval [0,t] is
{{{DELTA}}}{{{h=t(v[0]-gt/2)=v[0]t-gt^2/2}}} , so
{{{h=h[0]+v[0]t-gt^2/2}}}
The above equation can be used to find the answers to parts c) and d) .
 
If I can use {{{g="10 m / s / s"}}} ,
substituting that, along with {{{h[0]=0}}} and {{{v[0]=25}}} , i get
{{{h=10+25t-5t^2}}}
c) {{{h=10+25t-5t^2>=20}}}-->{{{10+25t-5t^2-20>=0}}}-->{{{-5t^2+25t-10>=0}}}-->{{{-t^2+5t-2>=0}}}
Since the solutions to {{{-t^2+5t-2=0}}} are {{{t=(5 +- sqrt(33))/2}}}
{{{-t^2+5t-2>=0}}}--->{{{(5 +- sqrt(33))/2<=t<=(5 +- sqrt(33))/2}}}
(An approximate decimal value may be expected as an answer).
d) {{{h=10+25t-5t^2=0}}}-->{{{-t^2+5t+2>=0}}}
The two solutions are {{{t=(-5 +- sqrt(33))/2}}} , but 
{{{t=(-5 - sqrt(33))/2<0}}} does not make sense,
so the rocket gets down to ground level at {{{t=(-5 + sqrt(33))/2}}} .
(An approximate decimal value may be expected as an answer).
If you must use {{{g=9.8}}}{{{m/s^2}}} ,
the calculations are the similar,
but I suggest using a calculator.