SOLUTION: how do solve three consecutive even integers? the question: find three consecutive even integers such that four times the square of the third, less three times the square of the f

Question 21301: how do solve three consecutive even integers?
the question: find three consecutive even integers such that four times the square of the third, less three times the square of the first, minus 41, is twice the square of the second..
my answer:4x^2=41
then thats where i get stuck can you help please.
Answer by Photonjohn(42) (Show Source): You can put this solution on YOUR website!
I didn't get your answer, but it appears it is not solvable.
Let X = 1st even integer
X+2 = next even integer
X=4 = 3rd even integer
4(X+4)^2 -3X^2 - 41 = 2(X+2)^2
4(X^2 + 8X + 16) - 3X^2 - 41 = 2(X^2 + 4X + 4)
4X^2 + 32X + 64 - 3X^2 - 41 = 2X^2 + 8X + 8
X^2 - 24X - 15 = 0
doesn't yield even integers
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