Question 414094
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Hi
Let x,(x+2),(x+4)represent the three consecutive positive even integers 
Question states***
  x^2 + (x+2)^2 + (x+4)^2 = 116
Solving for x
  3x^2 + 12x + 20 = 116
  3x^2 + 12x - 96 = 0
   x^2 + 4x - 32 = 0
factoring
 (x + 8)(x-4)=0  Note:SUM of the inner product(8x) and the outer product(-4x) = 4x
 (x + 8)=0  x = -8 is an Extraneous solution (not 'positive')
 (x-4)=0    x = 4  the three consecutive positive even integers are 4,6,8

CHECKING our Answer***
 16 + 36 + 64 = 116