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Seahawks' Adams signs 4-year extension worth reported $70M

Chris Coduto / Getty Images Sport / Getty

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 safeties

Player 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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