Seattle Seahawks and All-Pro safety Jamal Adams agreed to a four-year extension that can be worth up to $72 million. The deal, made official by the team on Tuesday, makes Adams the highest-paid player in his position with a $17.5 million annual salary.
The extension comes with a $38 million guarantee and a $20 million signing bonus. Adams is now under contract with the Seahawks through 2025.
The two sides were reportedly engaged in contract discussions for the better part of the offseason. Adams decided to report for training camp, deciding not to take the “holdout” route as the talks dragged out, but did not participate in the practices and team activities. Still, both parties were motivated to get the deal done and managed to do so before the season started.
“We officially signed. I’m excited to be here. It’s going to be a wonderful journey, man,” said Adams about the extension before adding that his next goal is to win a championship in Seattle.
Seahawks acquired Jamal Adams during the 2020 offseason, sending a trade package headlined by two first-round picks to the New York Jets. The deal immediately paid off for Seattle, as Adams had a monster season that saw him getting 9.5 sacks and breaking the NFL record for most sacks by a defensive back. He also had 59 tackles, three pass deflections, and one forced fumble en route to his third straight Pro Bowl nod and Second-team All-Pro selection.
