Former Texas Longhorns safety DeShon Elliott has signed a two-year extension worth $12.5 million that includes $9.21 million in guarantees with the Pittsburgh Steelers, NFL Network’s Tom Pelissero reports. The deal keeps him in Pittsburgh through the 2027 season.

Elliott, 28, had one of his best NFL seasons in 2024. In 15 games, 14 of which he started, the Dallas native had a career-high 108 total tackles, six passes defended, one interception, two forced fumbles and three fumble recoveries. In coverage, he allowed a passer rating of just 88.8 as the nearest defender, the lowest since his rookie season in 2019.

Last year, Elliott signed a two-year, $6 million deal with the Steelers in free agency. Now, he’s more than doubled his value in just one year.

Elliott entered the league as a sixth-round pick by the Baltimore Ravens in the 2018 NFL Draft, though he unfortunately missed his entire rookie season due to injury. After three seasons in Baltimore (not counting the aforementioned rookie season), he then spent the 2022 season with the Detroit Lions and the 2023 season with the Miami Dolphins, both on one-year contracts. Now, he’s seemingly found a long-term home for himself at last.

At Texas, Elliott became a star in the secondary as a junior in 2017. With a conference-leading six interceptions, two of which he returned for touchdowns, he earned first-team All-Big 12 and unanimous All-American honors.

Elliott may not be the biggest name to come out of Austin in recent years, but he has built a solid NFL career for himself, and should continue to excel in Pittsburgh for a while to come.