SEATTLE — The Seattle Seahawks have signed safety Jamal Adams to a four-year contract extension that is expected to make the former All-Pro the highest-paid safety in the NFL.
Adams was on the practice field for the first time this season after the deal was signed and announced by the team. NFL Network first reported the agreement, which is expected to include $38 million guaranteed and be worth up to $70 million.
Adams had been attending meetings but had yet to participate in on-field activities. Adams was acquired in a trade from the New York Jets last summer, a deal that cost Seattle two first-round picks.
Adams persevered through his first season with the Seahawks despite multiple injuries and received praise for his can-do attitude on and off the field.
"Adams brought something special to the Seahawks in his first season with the team," the Seahawks said in a press statement Tuesday. "It wasn't just the play-making ability, though there was plenty of that, but also the energy he brought to every practice and game, the competitiveness, the passion for the game and the leadership that showed in just about everything Adams did."
"Jamal had an incredible role in the season," Seahawks coach Pete Carroll said at the end of the season "The juice that he brought, the competitiveness, the attitude of toughness, you just couldn't miss it."
The Seahawks said agreeing on a contract extension with Adams ensures he'll be able to keep bringing all of those attributes to the team's defense for years to come.
