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.
Full details for the #Seahawks and QB Russell Wilson: 4 new years, $140M. $65M to sign. Total guarantee: $107M. A no-trade clause. ... Now, Wilson is under contract for 5 years and $157M. And staying in Seattle.— Ian Rapoport (@RapSheet) April 16, 2019
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.