Peterson will be joining New Orleans on a two-year deal with a $7 million base value.

After waiting more than a month, Adrian Peterson has found his next NFL landing spot.

The Minnesota Vikings’ all-time rushing leader finalized a deal Tuesday with the New Orleans Saints.

The deal will run two years and is worth $7 million with $3.5 million guaranteed, a person familiar with the contract told USA TODAY Sports’ Tom Pelissero. The pact’s maximum value with incentives and escalators is $15.25 million.

The person requested anonymity because the Saints had not revealed terms of the deal.

“I am excited to be joining the New Orleans Saints,” Peterson told ESPN’s Josina Anderson, who first reported his deal. “I’m really looking forward to this opportunity. Most importantly, I chose this team because it just felt right within my spirit. Additionally, my wife and family added their confirmation with the same feelings.”

USA TODAY Sports was among several outlets that confirmed Monday that Peterson and the Saints were close to an agreement.

The move brings an end to an extended free agency process for Peterson, 32, who rushed for 11,747 yards in 10 seasons. The Vikings chose not to pick up his $18 million option for 2017 and instead released him in March.

But he won’t have to wait long to face his former team. The Vikings host the Saints in the first Monday Night Football game of the upcoming season.


Peterson also visited the Seattle Seahawks and New England Patriots, but the teams opted to look elsewhere to address their needs in the backfield. The Seahawks signed former Green Bay Packers running back Eddie Lacy, while the Patriots took on Rex Burkhead and Mike Gillislee.

In New Orleans, Peterson will team up with Mark Ingram, who had a career-high 1,043 rushing yards last season.

PHOTOS: Notable NFL players on the move


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