SOLUTION: Find three consecutive multiples of 6 such that the sum of the squares of the the first two is 180 greater than the square of the third. I mostly need to know how to set this quest
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Question 469493: Find three consecutive multiples of 6 such that the sum of the squares of the the first two is 180 greater than the square of the third. I mostly need to know how to set this question up, but any help would be great. Answer by amoresroy(361) (Show Source):
You can put this solution on YOUR website! Find three consecutive multiples of 6 such that the sum of the squares of the the first two is 180 greater than the square of the third. I mostly need to know how to set this question up, but any help would be great.
Let 6x = the 1st number
6(x+1) = the 2nd number = 6x+6
6(x+2) = the 3rd number = 6x+12
(6x)^2 + (6x+6)^2 = (6x+12)^2 +180
Solve for x
