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

(x) ,  (x + 2)  consecutive odd integers
(x)^2 = 40 + (x + 2)^2 
x^2 = 40 + x^2 - 4x + 4
x^2 = x^2 - 4x + 44
subtract x^2 from both sides
0 = -4x + 44
subtract -4x from both sides
4x = 44
divide both sides by 4
x = 11* -1 = -11
x + 2 = -11 + 2 = -9
(-11)^2 = 40 + (-9)^2
121 = 40 + 81
121 = 121 CHECKS
-11 and -9 are the two negative integers.

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