1880 United States presidential election in Delaware explained

See main article: 1880 United States presidential election.

Election Name:1880 United States presidential election in Delaware
Country:Delaware
Flag Year:1880
Type:presidential
Ongoing:no
Previous Election:1876 United States presidential election in Delaware
Previous Year:1876
Next Election:1884 United States presidential election in Delaware
Next Year:1884
Election Date:November 2, 1880
Image1:WinfieldScottHancock2 (cropped 3x4).jpg
Nominee1:Winfield S. Hancock
Party1:Democratic Party (United States)
Home State1:Pennsylvania
Running Mate1:William H. English
Electoral Vote1:3
Popular Vote1:15,181
Percentage1:51.53%
Nominee2:James A. Garfield
Party2:Republican Party (United States)
Home State2:Ohio
Running Mate2:Chester A. Arthur
Electoral Vote2:0
Popular Vote2:14,148
Percentage2:48.03%
Map Size:210px
President
Before Election:Rutherford B. Hayes
Before Party:Republican Party (United States)
After Election:James Garfield
After Party:Republican Party (United States)

The 1880 United States presidential election in Delaware took place on November 2, 1880, as part of the 1880 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president.[1]

Delaware was won by General Winfield Scott Hancock (DPennsylvania), running with former Representative William Hayden English, with 51.53% of the popular vote, against Representative James A. Garfield (R-Ohio), running with the 10th chairman of the New York State Republican Executive Committee Chester A. Arthur, with 48.03% of the vote.[1]

As of 2023, this is the most recent time that a Republican won the presidency without carrying Sussex County.

See also

Notes and References

  1. Web site: 1880 Presidential Election Results Delaware.