SOLUTION: A number differs from it's square by the average of the number and it's square I sent the wrong email before

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Question 22674: A number differs from it's square by the average of the number and it's square
I sent the wrong email before
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let "x" be the number.
It's square is x^2
The average of the number and its square is (x+x^2)/2
Equation:
x-x^2 = (x+x^2)/2
2x -2x^2 = x+x^2
3x^2-x=0
x(3x-1)=0
x =0 or x = 1/3
Cheers,
Stan H.