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N² + 6N + 8 = 168
N² + 6N = 160
N = 10
(a+2)(a+4) = 168
a² + 4a +2a + 8 =168
a² + 6a +8 - 168 = 0
a² + 6a - 160 = 0
a 16
a -10
entonces
a + 16 =0 a - 10 = 0
a = -16 a =10
comprobamos
si a = 10
seria (a+2)(a+4) =(10+2)(10+4)=(12)(14)= 168 cumple
si a = -16
seria (a+2)(a+4) = (-16+2)(-16+4)= (-14)(-12)= 168 cumple
entonces esos números enteros pares consecutivos son 12, 14 y -12, -14