Question 953331
find three consecutive integers
x, (x+1), (x+2)
:
 such that three times the product of the first and last integers, decreased by 150, is equal to the square of the second integer.
3x(x+2) - 150 = (x+1)^2
distribute on the left, FOIL the right (x+1)(x+1)
3x^2 + 6x - 150 = x^2 +1x + 1x + 1
3x^2 + 6x - 150 = x^2 + 2x + 1; this looks like (b
collect like terms on the left
3x^2 - x^2 + 6x - 2x - 150 - 1 = 0
2x^2 + 4x - 151 = 0
The problem with this is the solutions are not integers
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