SOLUTION: The braking distance (in feet) of a car going V mph is given by {{{d(v)=v^2/20+v}}} v is greater or equal to 0. how fast would the car have been traveling for a braking distance

Question 332228: The braking distance (in feet) of a car going V mph is given by v is greater or equal to 0.
how fast would the car have been traveling for a braking distance of 150feet? round to nearest mile per hour.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
d(v)=v^2/20+v
(v^2/20)+v=150
v^2+20v=3000
v^2+20v-3000=0
v=46 mph Quadratic formula below
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Ed
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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=12400 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 45.6776436283002, -65.6776436283002. Here's your graph:

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