SOLUTION: An angry construction worker thros his wrench downward from a height of 128 feet with an initial vleocity of 32 feet per second, The height of the wrench above the ground after "t

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Question 416300: An angry construction worker thros his wrench downward from a height of 128 feet with an initial vleocity of 32 feet per second, The height of the wrench above the ground after "t" seconds is given by:
S(t) = -16r^2 -32t +128
(a) what is the height of the wrench afer 1 second? (this one I got. It's 80 feet)
WHAT I DON"T GET IS:
(b) How long does tit take for the wrench to reach the ground?
I'm guessing something along the line of: 0=-16t^2-32=128
t^2+2t-9=0
THEN I GET STUCK,,,, PLEASE HELP!!! THANKS
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
t^2+2t-9=0
Since you can't factor, you must resort to the "quadratic formula".
Doing so yields:
t = {2.162, -4.162}
You can throw out the negative solution leaving:
t = 2.12 seconds
.
Details of quadratic follows:
.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case 1t%5E2%2B2t%2B-9+=+0) has the following solutons:

t%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=%282%29%5E2-4%2A1%2A-9=40.

Discriminant d=40 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+40+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%282%29%2Bsqrt%28+40+%29%29%2F2%5C1+=+2.16227766016838
t%5B2%5D+=+%28-%282%29-sqrt%28+40+%29%29%2F2%5C1+=+-4.16227766016838

Quadratic expression 1t%5E2%2B2t%2B-9 can be factored:
1t%5E2%2B2t%2B-9+=+1%28t-2.16227766016838%29%2A%28t--4.16227766016838%29
Again, the answer is: 2.16227766016838, -4.16227766016838. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-9+%29