SOLUTION: When dealing with the quadratic equation f(x)=-0.2x^2+12x+11, where the function represents the expected number of ticket sales each day, and x=1 is the day tickets go on sale. M

Question 596129: When dealing with the quadratic equation f(x)=-0.2x^2+12x+11, where the function represents the expected number of ticket sales each day, and x=1 is the day tickets go on sale.
My main question is, using this equation how can I determine the last day that tickets will be sold?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
f(x)=-0.2x^2+12x+11
since f(x) represents the number of tickets sold
to find the last day tickets were sold, set f(x) to zero and solve for x:
f(x)=-0.2x^2+12x+11
0=-0.2x^2+12x+11
0=0.2x^2-12x-11
applying the quadratic formula, we get:
x = {60.9, -0.9}
we can toss out the negative solution (extraneous) leaving:
x = 60 days
.
details of quadratic follows:

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=152.8 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 60.9030742807249, -0.903074280724887. Here's your graph:

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