According to a person with knowledge of the negotiations who spoke with The Associated Press on Sunday night, the
nine-time Pro Bowl quarterback has decided to accept a one-year contract with the Steelers.
The individual, who spoke under anonymity since the deal hasn’t been finalized, stated that Wilson will get the veteran’s minimum salary of US$1.21 million, with the Denver Broncos covering the remaining US$39 million of his
compensation.
On X, the website that was once known as Twitter, Wilson announced his plans, writing: “Year 13. Thank you,
(at)Steelers.”
After being acquired by the Broncos in a trade with Seattle, Wilson, 35, had an 11-19 record in his first two seasons
there. After a terrible 2022 season, he rebounded to throw 3,070 yards, 26 touchdowns, and just eight interceptions
in his debut season under coach Sean Payton last year. Despite this, he lost his job to Jarrett Stidham.
Wilson spent ten seasons as the Seahawks’ quarterback, leading the team to eight postseason trips and a Super Bowl
victory.
Mason Rudolph was the starting quarterback for the Pittsburgh Steelers when they lost a wild-card playoff game.
After Kenny Pickett, who was 7-5 before suffering an injury, was replaced, Rudolph went 3-0. In the other two games, Mitch Trubisky started and finished 0–2.
This coming season, Pittsburgh is set to meet the Broncos in Denver; Wilson may make a return. It is anticipated
that the NFL will unveil its league schedule in May.
Wilson was granted permission to negotiate with other clubs, but the Broncos notified him last week that he will be
released on Wednesday, the start of the new league season.
“We thank Russell for his contributions and dedications to our team and community while wishing him the best as
he continues his career,” the team said in a tweet last week. It continued, “We are excited to improve this offseason
and will have the flexibility to get better through the draft and free agency.”
