Question 1205890
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The goal is to find the values of t which make h(t) = 0 true.


You can factor using trial and error, or use the quadratic formula.
I'll use the quadratic formula.


Plug in a = -1, b = 16, c = 80 
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-16+-sqrt((16)^2-4(-1)(80)))/(2(-1))}}}


{{{t = (-16+-sqrt(256 + 320))/(-2)}}}


{{{t = (-16+-sqrt(576))/(-2)}}}


{{{t = (-16+-  24)/(-2)}}}


{{{t = (-16+24)/(-2)}}} or {{{t = (-16-24)/(-2)}}}


{{{t = (8)/(-2)}}} or  {{{t = (-40)/(-2)}}}


{{{t = -4}}} or  {{{t = 20}}}


If t = -4 or t = 20, then h(t) = 0.
I'll let the student confirm this.


The leading coefficient being negative (a = -1) tells us that the parabola opens downward. Therefore, the portion of the curve above the x axis spans from t = -4 to t = 20
h(t) > 0 on the interval -4 < t < 20


{{{h(t) >= 0}}} on the interval {{{-4 <= t <= 20}}}


However, since t is a time value, it makes no sense to have t be negative.
So we must require that {{{t >= 0}}} as well.


Overlap {{{t >= 0}}} and {{{-4 <= t <= 20}}} together to find the final answer is {{{0 <= t <= 20}}}
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