Question

find two consecutive negative odd integers such that the square of the lesser integer is 40 more than the square of the greater integer

Answer 1

x^2 = (x + 2)^2 + 40
x^2 = x^2 + 4x + 4 + 40
x = -11

-11 and -9
121 - 81 = 40 correct

-11 and -9