The Seahawks and Jamal Adams have agreed to terms on a four-year, $70MM deal (Twitter link via NFL.com’s Ian Rapoport). The contract makes Adams the NFL’s highest-paid safety on a per-year basis with $38MM in guarantees.
The Seahawks traded for Adams last summer, sending two first-round picks, a third-round pick, and fellow safety Bradley McDougald to the Jets. The deal brought them a fourth-round pick in addition to one of the game’s most exciting young talents.
Adams – who won’t turn 26 until October – has been a certified star since his rookie season. Between 2018 and 2019, only Vikings standout Anthony Harris graded out higher than Adams at safety, according to Pro Football Focus. Eddie Jackson (Bears) was No. 3 during that stretch, and Adams has now leapfrogged him by a wide margin in earnings.
In 2019, Adams earned his first ever First-Team All-Pro nod while notching 75 tackles, seven passes defensed, 6.5 sacks, and a defensive touchdown. Last year, he missed four games with a groin injury, but still showed his mettle as a dangerous playmaker. He notched 9.5 sacks in just 12 games, the most any defensive back has ever posted in a single season. And, with 83 overall stops including eleven tackles for loss, Adams earned his third straight Pro Bowl nod.
Previously, Adams was set to enter his “walk year” and make less than $10MM. Of course, the Seahawks would have been able to franchise tag him next spring, which hurt his leverage somewhat. Ultimately, the two sides reached a logical compromise – Adams gets his mega-deal and a chance to cash in all over again during his prime. The Seahawks, meanwhile, managed to get him for slightly less than Bobby Wagner‘s $18MM-AAV deal.