SOLUTION: find 3 consecutive integers such that the square root of the first plus the product of the other two is 46.

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Question 258804: find 3 consecutive integers such that the square root of the first plus the product of the other two is 46.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The first integer has to be a square so that the square root is an integer.
+N=1 , sqrt%28N%29=1, %282%29%283%29=6, 1%2B6=7+
N=4, sqrt%28N%29=2, %285%29%286%29=30, 4%2B30=34
N=9, sqrt%28N%29=3, %2810%29%2811%29=110, 9%2B110=119
We're already over 46 and this function will only increase.
No solution.