Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.

N; n+1; n+2 - 3 consecutive numbers

n(n + 1) = (n + 2)² - 19    |use a(b + c) = ab + ac and (a + b)² = a² + 2ab + b²

n² + n = n² + 4n + 4 - 19    |subtract n² from both sides

n = 4n - 15    |subtract 4n from both sides

-3n = -15    |divide both sides by (-3)

n = 5

n + 1 = 5 + 1 = 6

n + 2 = 5 + 2 = 7

5; 6; 7.