The Seattle Seahawks and safety Jamal Adams have agreed on a four-year contract extension worth up to $72 million. Adams is now the highest-paid safety in the league and will make him $38 million in guaranteed money. There were concerns that the deal would not get done between the two sides, but Tuesday afternoon they came to an agreement. The Seahawks acquired Adams one year ago in a deal with the New York Jets that included a 2021 first-round pick, a 2022 first-round pick, a 2021 third-round pick, and safety Bradley McDougald.
The #Seahawks have agreed to terms on a large extension for star S Jamal Adams, a 4-year, $70M deal that makes him the league’s highest paid safety, I’m told. He gets $38M guaranteed, breaking the stalemate. A long time coming and well-deserved. ???
— Ian Rapoport (@RapSheet) August 17, 2021
Jamal Adams’ deal with the #Seahawks has a max value of $72 million over four years, meaning he could reach the $18 million per year plateau (i.e. Bobby Wagner’s average) with team and individual performance.
— Mike Garafolo (@MikeGarafolo) August 17, 2021
This deal was a necessity for the Seahawks who gave up several high-valued assets to acquire Adams last year. Last year with Seattle the talented safety posted 83 combined tackles with 59 of them being solo tackles. He also had a career-high 9.5 sacks and career-high 11 tackles for a loss. Injuries hampered him some last season and he only played 12 games in the regular season. Adams is still considered a top defensive back in the NFL despite dealing with some injuries last season.
Jamal Adams is now the NFL’s highest-paid safety
Throwback to where it all began @LSUfootball
(via @DraftRT)
— PFF College (@PFF_College) August 17, 2021
This deal secures the anchor of the Seattle secondary for at least the next few years. Seattle and Adams can now refocus on trying to win a Super Bowl.
