SOLUTION: 3 consecutive integers such that the square of the first plus the product of the other two is 191

SOLUTION: 3 consecutive integers such that the square of the first plus the product of the other two is 191

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Question 801544: 3 consecutive integers such that the square of the first plus the product of the other two is 191
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
consecutive integers: (x-1), x, (x+1)

(x-1)^2 + x(x+1) = 191
x^2 - 2x + 1 + x^2 + x = 191
2x^2 - x - 190 = 0
(x - 10)(2x + 19) = 0

x is an integer, so x = 10 and the integers are 9, 10, and 11.
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