1880 United States presidential election in Delaware

Main article: 1880 United States presidential election

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]

1880 United States presidential election in Delaware

← 1876 November 2, 1880 1884 →

Nominee Winfield S. Hancock James A. Garfield Party Democratic Republican Home state Pennsylvania Ohio Running mate William H. English Chester A. Arthur Electoral vote 3 0 Popular vote 15,181 14,148 Percentage 51.53% 48.03%

County Results Hancock   50–60% Garfield   50–60%

President before election Rutherford B. Hayes Republican Elected President James Garfield Republican

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]

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

Results

1880 United States presidential election in Delaware[1]
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%

See also

• United States presidential elections in Delaware