Need a formula; the product of two consecutive integers is 41 more than their sum. Find the integers.

Answer 1

let the two numbers be x and x+1

then x(x+1) = x + y + 41
expanding and simplifying we get
x^2 - x - 42 = 0
(x-7)(x+6) = 0
x = 7 or x = -6

case 1: the two numbers are 7 and 8
check: product = 56
sum = 15, 56 is greater than 15 by 41

case 2: the two numbers are -6 and -5
check: product is 30
sum is -11
30 is 41 greater than -11

so the two numbers are either 7 and 8 or
-6 and -5

Answer 2

Find such 3 consecutive integers that the sum of the first and third is 22 more than the second one.

Answer 3

×^-×-42=0