Question 373769
Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest. I would like to know how to solve because I already have the answer which is 6,7,8. 

Let x = the smallest integer
  x+1 = the next integer
 x+2  = the larget integer

(x+x+1)^2 = (x+2)^2 + 105
(2x+1)^2  = (x+2)^2 + 105
4x^2+4x+1 = x^2+4x+4+105
4x cancels out
4x^2-x^2  = 105+4-1
     3x^2 = 108
      x^2 = 36
        x = 6
     x+1  = 7
     x+2  = 8