The Seahawks have recently signed star safety Jamal Adams to a record-breaking deal. The deal is reportedly for four year, $70 million in total, and includes a $20 million signing bonus and $38 million guaranteed. The new contract will make Jamal Adams the highest-paid safety in the league.
Adams had previously reported to training camp. However, he did not practice or play at all as part of a holdout for a new contract. Adams is arguably the second-best defensive player on the Seahawks (behind LB Bobby Wagner), so a new deal was an important task this offseason. The Seahawks’ secondary is one of the weaker units in the league, and having their star safety on good terms will be a plus.
The standout safety had an impressive season last year. He broke the record for most sacks by a defensive back with 9.5. He totaled 83 tackles despite playing in 12 games. Adams was an outstanding pass rusher last season, totaling 14 tackles for loss and 30 pressures. That’s an impressive feat for being a safety.
Adams faced a fair amount of backlash from the media this year, as he failed to record an interception. Some Seahawks fans may have been disappointed by this given Seattle traded quite a few first-round picks for him.
The Seahawks sent a massive amount of assets to the New York Jets to acquire Adams last summer. The trade included safety Bradley McDougald, a first-round pick in 2021 and 2022, and a third-round pick in 2021.
The former LSU product formed a strong duo with free safety Quandre Diggs in the Seahawks secondary. The two will both look to take Seattle’s mediocre defense to the next level. This was the reasoning behind the trade for Adams in the first place.
Adams struggled with injuries this season as he missed 4 games due to a torn labrum, two broken fingers, and a hyperextended elbow.