SOLUTION: Find three consecutive integers such that the product of the second and the third is 16 less than the square of the first. I'm not quite sure how to set up this equation, and I

Question 959752: Find three consecutive integers such that the product of the second and the third is 16 less than the square of the first.
I'm not quite sure how to set up this equation, and I direly require assistance.
I made an attempt, and I came up with: x^2-16(x+1)(x+2).
However, I am almost certain this is incorrect.
Please, please help me.
-Terrible Algebra Student
Answer by JoelSchwartz(130) (Show Source): You can put this solution on YOUR website!
x=first number
y=second number
z=third number
y=x+1
z=x+2
yz=x^2-16
(x+1)(x+2)=x^2-16
x^2+x+2x+2=x^2-16
x^2+3x+2=x^2-16
3x+2=-16
3x=-18
x=-6
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