Jamal Adams, Seahawks Agree to Reported $70M Contract; Is NFL's Top-Paid Safety

Jamal Adams will remain with the Seattle Seahawks after the two sides agreed to a four-year extension on Tuesday. 

Per NFL Network's Ian Rapoport, Adams will become the highest-paid safety in the NFL after agreeing to a four-year, $70 million deal that includes $38 million guaranteed. 

Adam Schefter of ESPN noted Adams' new deal can max out at $72 million.

The 25-year-old is coming off a huge first season with the Seahawks, tallying 9.5 sacks in just 12 games to earn his third straight Pro Bowl selection.

Adams provides a unique ability to rush the passer from the secondary that is unmatched among those at his position:

The 2016 No. 6 overall draft pick was set to hit free agency after the 2021 season, but it was important for Seattle to lock him down after giving up two first-round picks in a trade with the New York Jets. The squad also gave up safety Bradley McDougald and a 2021 third-round pick in exchange for the strong safety and a 2022 fourth-round pick.

The LSU product had reportedly requested a trade from New York as he was seeking a new contract with the team worth more than $20 million per year, per Jeremy Fowler of ESPN. Gregg Bell of the News Tribune reported the Seahawks expected a deal worth over $16 million per year since making the trade. Adams was voted first-team All-Pro in 2019 after totaling 6.5 sacks and two defensive touchdowns.

The Texas native ranked third on the team with 83 tackles and helped the squad finish 15th in points allowed en route to a 12-4 record.

The long-term deal gets Seattle more value from its original trade while locking down an impactful player on the defensive side of the ball.