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 (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.