1880 United States presidential election in Delaware
The 1880 United States presidential election in Delaware took place on November 2, 1880, as part of the 1880 United States presidential election. Delaware voters chose 3 representatives, or electors, to the Electoral College, who voted for president and vice president.[1]
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Elections in Delaware | ||||||||||||
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Delaware was won by General Winfield Scott Hancock (D–Pennsylvania), 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]
Results
United States presidential election in Delaware, 1880[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 15,181 | 51.53% | 3 | 100.00% | ||
Republican | James A. Garfield of Ohio | Chester A. Arthur of New York | 14,148 | 48.03% | 0 | 0.00% | ||
Greenback | James B. Weaver of Iowa | Barzillai J. Chambers of Texas | 129 | 0.44% | 0 | 0.00% | ||
Total | 18,345 | 100.00% | 3 | 100.00% |
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