SOLUTION: If an object is propelled upward from a height of 96 feet at an initial velocity of 80 feet per second,then its height after t seconds is given by the equation h= -16t^2+80t+96 whe

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Question 315125: If an object is propelled upward from a height of 96 feet at an initial velocity of 80 feet per second,then its height after t seconds is given by the equation h= -16t^2+80t+96 where h is in feet. After how many seconds will the object reach a height of 196 feet.
Answer by malaydassharma(59) About Me  (Show Source):
You can put this solution on YOUR website!
-16t%5E2%2B80t%2B96+=h
OR -16t%5E2%2B80t%2B96+=196
OR 16t%5E2-80t%2B100+=0

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 16x%5E2%2B-80x%2B100+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-80%29%5E2-4%2A16%2A100=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%28-80%29%29%2F2%5C16.
Expression can be factored: 16x%5E2%2B-80x%2B100+=+16%28x-2.5%29%2A%28x-2.5%29

Again, the answer is: 2.5, 2.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+16%2Ax%5E2%2B-80%2Ax%2B100+%29