CLEVELAND - The Cleveland Indians have made a one-year, $17.4 million qualifying offer to first baseman Carlos Santana.
The team announced the offer Monday afternoon. MLB.com's Jordan Bastian reported the specifics of the offer.
Indians make it official: Santana gets 1-year, $17.4M Qualifying Offer. Has 10 days to accept/decline. More on QO: https://t.co/I65J5SeVs9— Jordan Bastian (@MLBastian) November 6, 2017
Santana has 10 days to accept the team's offer.
“I’m hopeful,” Santana said after the Indians fell to the New York Yankees in the American League Division Series. “I’m hopeful that I come back. This is my house. Everybody knows me."
During the 2017 season, Santana hit .259 with 23 home runs, 37 doubles, 79 runs batted in, another 90 scored and 148 hits over 154 games. Additionally, Santana struck out 94 times, but worked his way to 88 walks.
In his eight seasons with the Indians, Santana has crafted a .249 batting average with 995 hits in 3,994 at-bats over 1,116 regular-season games. He belted 236 doubles, 13 triples and 174 home runs, drove in 587 runs, scored another 573 and drew 726 walks against 812 strikeouts.
Originally a catcher when promoted to the Major League club during the 2010 season, Santana has played behind the plate, at third base, in left field and settled into a starter’s role at first base over the last four years.
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