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Megill closed the door on his 19th save Sunday, allowing two runs on two hits while striking out one over one inning against the Twins.

Megill entered the ninth with a three-run lead but managed to make things interesting, allowing a leadoff homer to Byron Buxton followed by a Max Kepler double. While Kepler eventually came around to score on a groundout, Megill retired the final three batters to finish off his 19th save. Sunday marked the second time in three appearances that Megill has allowed two runs after having allowed no more than one earned run during any of his previous 31 appearances.

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