SOLUTION: find 3 consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 7 times the third

Question 428350: find 3 consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 7 times the third
Answer by sudhanshu_kmr(1152) (Show Source): You can put this solution on YOUR website!

three consecutive integers are x, x+1 and x+2.

x*(x+2) - (x+1) = 1+ 7(x+2)
=> x^2 + 2x - x -1 = 1 + 7x + 14
=> x^2 -6x -16 = 0
=> x^2 -8x + 2x -16 = 0
=> x(x-8) +2(x -8) = 0
=> (x+2)(x-8) =0
so, x=8 and x = -2 (not positive integer)
numbers are 8, 9 and 10.

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