Aaron Rodgers Signs 1-Year Deal Worth up to $19.5 Million, Will Wear #8 in Pittsburgh

That Aaron Rodgers was headed to Pittsburgh was one of the worst-kept secrets in football. All that remained a mystery were the details. Now, we know at least two key details.
Rodgers signed a one-year contract with Pittsburgh that will pay him a base salary of $13.65 million and $10 million guaranteed. However, when incentives are included, the deal has a maximum value of $19.5 million.
While the deal appears lucrative enough, it makes Rodgers the second-lowest-paid NFL quarterback not on a rookie contract, according to Pro Football Talk.
The other bit of news learned this week is that Rodgers will wear #8 in Pittsburgh, the same number he wore when playing for the Jets in New York. Rodgers, of course, wore the #12 for the first 18 years of his career in Green Bay. The move to the Jets brought a change in number, as the Jets had already retired the #12 in honor of the legendary Joe Namath. The Steelers also have a famous quarterback who wore the #12, four-time Super Bowl champion, Terry Bradshaw.
However, while the Steelers have not officially retired Bradshaw’s jersey, the team has not given another player that number in honor of Bradshaw. And they’re not going to do it for the future Hall of Famer Aaron Rodgers either.
Add to that the fact that Bradshaw has voiced his strong opposition to the Steelers signing Rodgers. It just wasn’t going to happen.
Bradshaw isn’t the only one underwhelmed by Rodgers wearing black and yellow next year.
DraftKings only moved the Steelers from +4500 odds to win the Super Bowl to +4000. Not nothing, but hardly a significant shift. Additionally, BetMGM didn’t adjust the Steelers’ Super Bowl odds at all, leaving Pittsburgh at 45:1.