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Verified Answer :

\(2n(2n+2)=168\\ 4n^2+4n-168=0\\ n^2+n-42=0\\ n^2+7n-6n-42=0\\ n(n+7)-6(n+7(=0\\ (n-6)(n+7)=0\\ n=6 \vee n=-7\\ -7 ot \in \mayhbb{N}\\\\ 2n=12\\ 2n+2=14\)

These numbers are 12 and 14.