Answers

2014-10-27T21:05:16+08:00
Let 'x' be one of the numbers
    'x+1' be the second number
   'x+2' the last number
squaring the three numbers you'll have:
x² + (x+1)² + (x+2)² = 434
x² + x² + 2x + 1 + x² + 4x + 4 = 434
3x² + 6x + 5 - 434 = 0
3x² + 6x - 429 = 0
dividing the whole equation with 3
x² + 2x - 143 = 0
(x + 13)(x-11) = 0
x = -13, x = 11
taking the negative integer you'll have the other 2 as:
x = -13
x + 1 = -13 + 1
         = -12
x + 2 = -13 + 2
        = -11
therefore the numbers are -13,-12,-11
0