- Oops!Something went wrong.Please try again later.
The chess match between the Seattle Seahawks and safety Jamal Adams has come to an end. After so many discussions involving the former LSU safety, he and the team have agreed on a four-year extension that makes him the highest-paid safety in the NFL.
Adams is going into his second season with the Seahawks after being traded from the New York Jets prior to the 2020 season. Adams hadn’t participated in training camp as he was trying to get a long-term deal done. Some reports stated Seattle was prepared to use the franchise tag after the season.
All of that is null and void with Tuesday’s news of the extension.
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
With Adams all locked up beyond this year, he can focus on the upcoming season and pursuing a Lombardi Trophy. Since his debut in the NFL, the former Tiger has appeared and started 58 games. He played his fewest last season due to injuries that required offseason surgery.
Adams has two career interceptions with one for a touchdown. He scored another touchdown on one of his four career fumble recoveries. Adams also has seven forced fumbles, 28 passes defended, 356 tackles, and 21.5 sacks to his credit in four seasons.