Seahawks reportedly make Jamal Adams highest-paid safety with 4-year, $70 million deal

·2-min read

The Seattle Seahawks and Jamal Adams reached a deal on a major contract extension. According to NFL Network's Ian Rapoport, Adams will sign a four-year, $70 million deal that will make him the highest-paid safety in the NFL.

This is what the list of highest-paid safeties looks like now, including the Adams deal.

Risky trade paying off for Seattle

Adams had been with the New York Jets his entire career, which started when they took him with the sixth overall pick in the 2017 draft up until July 2020. That's when the Seahawks traded for him in a massive deal that cost Seattle safety Bradley McDougald, first-round and third-round picks in the 2021 draft, and a first-round pick in the 2022 draft. 

That's a lot to give up, especially since Adams hadn't yet signed a long-term deal, but he was worth it. He was a star whenever he was on the field, setting the single-season record for sacks by a defensive back with 9.5. That's more than impressive, considering he did it in 12 games while playing through a host of injuries. A groin injury is what caused him to miss four games, but he also had injuries to both shoulders, two broken fingers and a hyperextended elbow. He played through the majority of that, and ended up having surgery on his fingers and one shoulder during the offseason. 

According to Rapoport, Adams was sitting out of practice while the deal was being negotiated. With that out of the way, he can get back to work a well-paid man. 

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