SOLUTION: Find three consecutive positive odd integers such that twice the square of the first one is one less then the product of the second and third

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Question 1053784: Find three consecutive positive odd integers such that twice the square of the first one is one less then the product of the second and third
Answer by jorel555(1290) About Me  (Show Source):
You can put this solution on YOUR website!
Let n be the first odd integer. Then the next integers will be n+2 and n+4. So:
2nē+1=(n+2)(n+4)
2nē+1=nē+6n+8
nē-6n-7=0
(n-7)(n+1)=0
n=7, -1
Your first integer is 7; the next two are 9 and 11. ☺☺☺☺