Question 1065741
There is a missing minus sign.
{{{h}}}{{{"="}}}{{{-16t^2+96t}}} is the height (in feet) of the object {{{t}}} seconds after it is thrown up.
A graph of {{{h}}} as a function of {{{t}}} looks like this
{{{graph(400,400,-2,8,-12,156,(-16x^2+96x)sqrt(6x-x^2)/sqrt(6x-x^2))}}}
When {{{h=144}}}, we have
{{{144=-16t^2+96t}}} .
Solving that equation for {{{t}}} we can find
when the object will reach a height of 144 feet 
(as {{{t}}} , in seconds after being thrown).
{{{-16t^2+96t=144}}}
Dividing both sides of the equal sign by {{{-16}}} ,
we get the equivalent equation
{{{t^2-6t=-9}}}
That can be solved by "completing the square":
{{{t^2-6t+9=-9+9}}}
{{{t^2+6t+9=0}}}
{{{(t-3)^2=0}}}
So, the only solution is {{{highlight(t=3)}}} .
The object is at a height of 144 ft only once,
{{{highlight(3seconds)}}} after being thrown up.
That means that it takes the object {{{highlight(3seconds)}}} to reach a height of 144 feet.
 
It also means that {{{144ft}}} is the maximum height.
After {{{3seconds}}} the object is falling back down.
Quadratic functions, like {{{h}}}{{{"="}}}{{{-16t^2+96t}}} ,
are symmetrical,
so if it takes the object {{{3}}} seconds to go
from {{{h=0}}} at {{{t=0}}} to {{{h=144}}} ,
it will take another {{{highlight(3seconds)}}} for the object to return
from {{{h=144}}} to {{{h=0}}} .
That is also obvious from the equation.
{{{h}}}{{{"="}}}{{{-16t^2+96t}}}{{{"="}}}{{{-16t(t-6)}}}
is zero obviously for {{{t=0}}} and {{{t=6}}} .
So after reaching a maximum height of 144 feet at 3 seconds,
it takes the object {{{6-3=3}}} seconds
to come back down to the ground (to {{{t=0}}} ).