Seattle - Seattle Seahawks quarterback Russell Wilson is to become the highest-paid player in the NFL after agreeing a four-year contract extension worth $140 million (R2 billion), reports said on Tuesday.
Wilson appeared to confirm the reports with a video message on Twitter.
"Seattle, we got a deal," Wilson said.
"Go Hawks. But I'ma see y'all in the morning. Time for y'all to go to bed."
ESPN reported that the new contract would tie the 30-year-old quarterback to the Seahawks through the 2023 season.
The deal includes a $65 million signing bonus and $107 million in guaranteed money.
With an average salary of $35 million per year, Wilson's deal eclipses Green Bay Packers quarterback Aaron Rodgers' deal signed last year which was then the richest contract in the NFL.
Rodgers salary is worth around $33.5 million a season. Wilson's signing bonus also tops Rodgers' then-record signing fee of $57.5 million.
Wilson's deal also includes a no-trade clause, ESPN reported.
The deal comes after an uneasy standoff between Wilson and the Seahawks.
Wilson, a five-time Pro Bowler who led the Seahawks to their one and only Super Bowl victory in 2014, had given the team a deadline of midnight on Monday to get a new deal put in place. This year marks the final year of a contract extension agreed in 2015 which would have seen him paid $17 million in 2019.
Wilson however had always indicated his preference to remain in Seattle, telling reporters at the end of last season he wanted to stay.
"It's business and everything," he said.
"I know essentially after the season, I could potentially be a free agent, that kind of thing. I don't think that way. I see myself being in Seattle. I love Seattle, and it's a special place for me."
Wilson's deal comes after the most statistically successful season of his career in Seattle. In 2018, he achieved a 65.6% completion rate for passing, while throwing for 35 touchdowns, his highest ever numbers.