Question 1140923
{{{h(t)=-6t^2+32t+3}}}

here you have downward parabola, and its maximum is vertex ({{{h}}},{{{k}}})

so, rewrite your equation in vertex form 

{{{h(t)=a(t-h)^2+k}}}

{{{h(t)=(-6t^2+32t)+3}}}

{{{h(t)=6(t^2-(16/3)t)+3}}}

{{{h(t)=6(t^2-(16/3)t+b^2)-(-6)b^2+3}}}........{{{b=(16/3)/2=16/6=8/3}}}

{{{h(t)=6(t-8/3)^2+6(8/3)^2+3}}}

{{{h(t)=6(t-8/3)^2+6(64/9)+3}}}

{{{h(t)=6(t-8/3)^2+128/3+3}}}

{{{h(t)=6(t-8/3)^2+128/3+9/3}}}

{{{h(t)=6(t-8/3)^2+137/3}}}

vertex is at ({{{8/3}}},{{{137/3}}})

the rocket to reach it's maximum height in {{{8/3}}} or {{{2.7}}} seconds