Question 1057459
 Find three consecutive negative integers such that the product of the first two is 26 greater than the square of the third.
let our numbers be:(n-2), (n-1), n
(n-2)(n-1) = n^2 + 26
FOIL the left
n^2 - 3n + 2 = n^2 + 26
combine like terms
n^2 - n^2 - 3n = 26 - 2
-3n = 24
n = 24/-3
n = -8
:
the three number -10, -9, -8
:
:
see if the works
-10 *-9 = -8^2 + 26
+90 = +64 + 26