SOLUTION: The height of an object thrown in the air is given by h = t2 � 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest

Question 224402: The height of an object thrown in the air is given by h = t2 � 7t + 3, where h is in feet and t is the time in seconds the object has been in motion. At what time (to the nearest tenth) is the object 10 feet in the air?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
h = t^2 � 7t + 3
Set h to zero and solve for t:
0 = t^2 � 7t + 3
Since we can't factor, we must resort to the quadratic equation. Doing so, yields:
x = {6.5, 0.5}
So, the object will be 10 feet at 0.5 seconds and 6.5 seconds.
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Details of quadratic equation follows...
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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=37 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6.54138126514911, 0.45861873485089. Here's your graph:

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