ANSWER:-
Let x and (x+2) be the consecutive even integers. We get
x 2+(x+2) 2 =100
expanding, we get x2 +x 2+4x+4=100
setting = 0, we get 2x 2+4x−96=0
Dividing by 2 , we get, x 2+2x−48=0
factoring, we get,
(x+8)(x−6)=0
solving for x, we get
x+8=0 = x=−8
and x−6=0=x=6.
so, we have
(−8,−6) OR (6,8)