The 1864 United States presidential election in Delaware took place on November 8, 1864, as part of the 1864 United States presidential election. State voters chose three representatives, or electors, to the Electoral College, who voted for president and vice president.[1]
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 County Results
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Delaware was won by the Democratic nominee, 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton. They defeated the National Union nominee, incumbent President Abraham Lincoln of Illinois and his running mate Senator and Military Governor of Tennessee Andrew Johnson.[1] McClellan won the state by a margin of 3.62%.
With 51.81% of the popular vote, Delaware would prove to be McClellan's third strongest state after Kentucky and New Jersey, his only two other winning states.[2]
Results
edit| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | George B. McClellan | 8,767 | 51.81% | |
| National Union | Abraham Lincoln (incumbent) | 8,155 | 48.19% | |
| Total votes | 16,922 | 100.00% | ||
See also
editReferences
edit- ^ a b c "1864 Presidential Election Results Delaware".
- ^ "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.