1868 United States presidential election in Delaware explained

See main article: 1868 United States presidential election.

Election Name:1868 United States presidential election in Delaware
Country:Delaware
Type:presidential
Ongoing:no
Previous Election:1864 United States presidential election in Delaware
Previous Year:1864
Next Election:1872 United States presidential election in Delaware
Next Year:1872
Election Date:November 3, 1868
Image1:Hon. Horatio Seymour, N.Y - NARA - 528568 (cropped).jpg
Nominee1:Horatio Seymour
Party1:Democratic Party (United States)
Home State1:New York
Running Mate1:Francis Preston Blair, Jr.
Electoral Vote1:3
Popular Vote1:10,957
Percentage1:59.00%
Nominee2:Ulysses S. Grant
Party2:Republican Party (United States)
Home State2:Illinois
Running Mate2:Schuyler Colfax
Electoral Vote2:0
Popular Vote2:7,614
Percentage2:41.00%
Map Size:210px
President
Before Election:Andrew Johnson
Before Party:Democratic Party (United States)
After Election:Ulysses S. Grant
After Party:Republican Party (United States)

The 1868 United States presidential election in Delaware took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Delaware voted for the Democratic nominee, Horatio Seymour, over the Republican nominee, Ulysses S. Grant. Seymour won the state by a margin of 18%.

With 59% of the popular vote, Delaware would prove to be Seymour's fifth strongest state in terms of popular vote percentage after Kentucky, Louisiana, Maryland and Georgia.[1]

See also

Notes and References

  1. Web site: 1868 Presidential Election Statistics. Dave Leip’s Atlas of U.S. Presidential Elections. 2018-03-05.