HELP WITH PRACTICE ACT MATH QUESTION: the braking distance, y feet, for Damon's car to come to a complete stop is modeled by y=3(x^2+10x)/40, where x is the speed of the car in miles per hour. According to this model, which of the following is the maximum speed, in miles per hour, Damon can be driving so that the braking distance is less than or equal to 150 feet?

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HELP WITH PRACTICE ACT MATH QUESTION: the braking distance, y feet, for Damon's car to come to a complete stop is modeled by y=3(x^2+10x)/40, where x is the speed of the car in miles per hour. According to this model, which of the following is the maximum speed, in miles per hour, Damon can be driving so that the braking distance is less than or equal to 150 feet?

Mathematics
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Have you considered solving: 150=3(x^2+10x)/40
no, i am not really trying to solve the problem as much as to know the steps TO solve it!:-) I'm practicing for the ACT

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for the final step of the equation would it be 6,000+33x^3?
Well, then that is what you will have to do. You should break out your best Quadratic Equation solver that you have stored in your head.
i am just learning algebra 2 this year, math is my weakest subject
could you help me with the steps of solving that equation please?
because when i do it all i get 6000+3x^2+30x and i do not know how to solve that. Could you please explain
I might do this: 150=3(x^2+10x)/40 50=(x^2+10x)/40 2000=(x^2+10x) x^2 + 10x - 2000 = 0 10^2 - 4(1)(-2000) = 100 + 8000 = 8100 sqrt(9100) = 90 Thus, (-10 + 90)/2 = 80/2 = 40 and we are done. No need to worry about (-10 - 90)/2 because that is negative and doesn't concern us in this problem.
thank you so much! my ACT scores thank you too:-)))) @tkhunny
@tkhunny but on the second step how did you get 50?
Look carefully. Division by 3.

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