SOLUTION: Find three consecutive positive integers such that the product of the first and second is 37 less than the square of the third.

Question 1080387: Find three consecutive positive integers such that the product of the first and second is 37 less than the square of the third.
Answer by rapture(86) (Show Source): You can put this solution on YOUR website!
Let: x, (x + 1), and (x + 2) be the three consecutive positive integers.

Here is the set up:

x(x + 1) = (x + 2)^2 - 37

Solve the equation for x. After finding x, go back and substitute the x-value into

x, (x + 1), and (x + 2) to find your three consecutive integers.
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