Find three positive consecutive integers such that the product of the first and second is two more than three times the third. Write and square a quadratic equation to find the three integers.

Answer 1

Let them be x-1, x, and x+1

(x-1)x = 2+3(x+1)

x^2 -x = 2 +3x +3

x^2 -4x -5 = 0

(x-5)(x+1) = 0

x=5, x=-1

-2, -1, 0 are 3 consecutive integers such that (-1)(-2)=2+3(0)

also

4,5,6 are consecutive positive integers such that 4x5 = 20 = 2+3x6 = 2+18 =20

You want the positive integers, so the only solution is 4, 5 and 6