The deal is done.
The Seahawks and safety Jamal Adams have agreed on a new four-year contract extension that will make Adams the highest paid safety in the NFL, according to multiple reports.
The deal was first reported by Ian Rapoport of the NFL Network.
The new contract for Adams is worth $70 million over four years with $38 million guaranteed. The numbers match those reported by Bob Condotta and Adam Jude of The Seattle Times last week as the standing offer to Adams that the team claimed was their final offer to the star safety. The deal brings an average of $17.5 million a year for Adams, which blows the ceiling off the safety market. Denver Broncos safety Justin Simmons had previously been the benchmark for the position at an average of $15.3 million a season.
Per Mike Garafolo of the NFL Network, the total value of the deal is up to $72 million with included incentives. The $18 million a year threshold is what linebacker Bobby Wagner makes on his deal with the team and it appears Seattle did not want Adams deal to surpass that of Wagner. It will take Adams reaching all of the incentives in the deal for him to match that of Wagner's contract.
Adams, and left tackle Duane Brown, have been sitting out through the first three weeks of training camp as they've sought new contracts with the team. While a deal for Brown has seemingly felt unlikely, it felt inevitable that a deal would get down between the Seahawks and Adams.
Photo Credit: PHILADELPHIA, PENNSYLVANIA - NOVEMBER 30: Jamal Adams #33 of the Seattle Seahawks celebrates a stop Philadelphia Eagles during the first quarter at Lincoln Financial Field on November 30, 2020 in Philadelphia, Pennsylvania. (Photo by Elsa/Getty Images)