<PRE><B>Question 119712</B>
Find three consecutive integers such that the sum of their squares is 77.
1st: x
2nd: x+1
3rd: x+2
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EQUATION:
x^2 + (x+1)^2 + (x+2)^2 = 77
3x^2 +6x +5 = 77
3x^2+6x-72=0
x^2+2x-24=0
(x+6)(x-4)=0
x = -6 or x=4
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If x = -6, then x+1 = -5, and x+2 = -4
If x = 4, then x+1 = 5, and x+2 = 6
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Cheers,
Stan H.</PRE>