Seahawks' Adams signs 4-year extension worth reported $70M
Seattle Seahawks safety Jamal Adams signed a four-year extension Tuesday, tying him to the club through the 2025 season, the team announced.
Adams secured a $70-million deal that includes $38 million guaranteed and makes him the highest-paid player at his position, reports NFL Network's Ian Rapoport.
The agreement came after extension talks reportedly hit a wall. The Seahawks were apparently prepared to use the franchise tag on Adams in 2022 once his rookie deal expired; he reportedly would have tried to secure a more lucrative tag by arguing he plays linebacker instead of safety.
Seattle acquired Adams from the New York Jets last offseason in a blockbuster trade. The Seahawks gave up two first-round picks and a third-round selection in the deal.
After starting slowly with the Seahawks by his standards, Adams rounded into form down the stretch last season. The 25-year-old finished the campaign with a team-high 9.5 sacks, blitzing more frequently than he did with the Jets. He underwent offseason shoulder and finger surgeries to repair injuries that may have contributed to a regression in coverage.
The Seahawks have now reset the market at three different positions over the last three years. In 2019, they handed Russell Wilson a quarterback record of $35 million per year and signed Bobby Wagner to a linebacker record of $18 million per season.
NFL's highest-paid safetiesPlayer | Team | AAV |
---|---|---|
Jamal Adams | Seahawks | $17.5M |
Justin Simmons | Broncos | $15.25M |
Budda Baker | Cardinals | $14.75M |
Eddie Jackson | Bears | $14.6M |
Kevin Byard | Titans | $14.1M |
A three-time Pro Bowler, Adams reported to Seahawks training camp on time but sat out practices while awaiting a contract resolution. He'll return to a defense that surrendered 285 passing yards per game last season, which ranked second-worst in the NFL.
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