The Seattle Seahawks have made Jamal Adams the highest-paid safety in the NFL, after signing him to a four-year extension, NFL Network Insider Ian Rapoport reported, Tuesday. Adams confirmed with a celebratory video posted on the team’s Twitter feed.
The deal contains a maximum value of $72 million, including a $20 million signing bonus and $38 million in guaranteed earnings. The extension is added to the final year of Adams’ rookie deal. He is now under contract through the 2025 season.
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Adams is now the Seahawks’ third-highest-paid player after quarterback Russell Wilson, who earns $35 million per season, and linebacker Bobby Wagner, who earns $18 million per season.
“We officially signed. I’m excited to be here,” Adams said in the video posted to Twitter. “It’s going to be a wonderful journey, man. The next thing on our mind is getting that [championship] — getting right back to it and going to get it.”
In his first four years in the NFL, he set the league’s single-season record for sacks by a defensive back last season with 9.5. Adams has also been named to three Pro-Bowl teams and is a one-time First-team All-Pro.
The Seahawks will take on the Indianapolis Colts in Week 1, next month.
Check out Adam’s announcement below.
[Via]