For the last few weeks there was genuine speculation that Russell Wilson would be suiting up somewhere other than Seattle for the 2019-2020 season as the Seahawks were appearing stubborn in regards to extending their Super Bowl-winning quarterback. 


On Monday, it was reported that ​the two sides were reportedly working on an extension as Wilson's self-appointed deadline approached, hinting that the franchise was beginning to come to its senses. And on early Tuesday morning, a deal was completed, making Russell the highest-paid player in the NFL.  


He and the Seahawks reportedly agreed to a four-year, $140 million contract extension, which includes a $65 million signing bonus, keeping him in Seattle for five more seasons.

Per NFL Network's Ian Rapoport, $107 million is guaranteed and Wilson has a no-trade clause.


The five-time Pro Bowler is now technically under contract with the Seahawks for five years and will make $157 million over that span. That's good for an AAV of $31.4 million. 

In the final season of the four-year, $87.6 million extension he signed in the summer of 2015, Wilson was set to make just $17 million in 2019. 


​Russell demanded the money and simply got what he deserved for giving Seattle seven years of sheer excellence.