Maggie wrote

Write three consecutive positive integers such that twice the product of the first and third is 2 less than 12 times the second.

x = 1st

x + 1 = 2nd

x + 2 = 3rd

2[x(x + 2)] = 12(x + 1) - 2 ←twice the product of 1st and 3rd is 2 less than 12 times 2nd

2(x² + 2x) = 12x + 12 - 2 ← distributive property

2x² + 4x = 12x + 10 ← distributive property

2x² - 8x - 10 = 0 ← subtracted 12x and 10

2(x² - 4x - 5) = 0 ← factored 2 out

2(x - 5)(x + 1) = 0 ← factored into two binomials

(x - 5)(x + 1) = 0 ← divided by 2

x - 5 = 0 or x + 1 = 0 ← set each factor equal to zero

x = 5 or x = -1 ← added 5 and subtracted 1

x + 1 = 6

x + 2 = 7

**5,6,7** are the 3 consecutive positive integers

- Proof