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Let x and (x+2) represent the two consecutive 'positive' even integers
Question states***
x^2 + (x+2)^2 = 164
Solving for x
x^2 + x^2 + 4x + 4 = 164
2x^2 + 4x - 160 = 0
x^2 + 2x - 80 = 0
factoring
(x+10)(x-8)= 0 Note:SUM of the inner product(10x) and the outer product(-8x)=2x
(x+10)=0 x = -10 Tossing out negative solution for 'positive integer'
(x-8)= 0 x = 8 the two consecutive 'positive' even integers are 8,10
CHECKING our Answer***
64 + 100 = 164