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.
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]