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.

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