SOLUTION: Divide 20 into two parts such thay three times the square of one part exceeds the other part bye 10

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Question 865482: Divide 20 into two parts such thay three times the square of one part exceeds the other part bye 10
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Let x and (20-x) represent the two parts

    3x^2 -10 = (20-x)

    3x^2 + x -30 = 0

    (3x +10)(x-3) = 0,  x =  3      (Tossing out the negative solution for unit measure)

And...Checking

   27 -10 = 20-3 = 17