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)   (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
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