See main article: 1864 United States presidential election.
| Election Name: | 1864 United States presidential election in Delaware |
| Country: | Delaware |
| Flag Year: | 1846 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1860 United States presidential election in Delaware |
| Previous Year: | 1860 |
| Next Election: | 1868 United States presidential election in Delaware |
| Next Year: | 1868 |
| Election Date: | November 8, 1864 |
| Image1: | GeorgeMcClellan2 (cropped).jpg |
| Nominee1: | George B. McClellan |
| Party1: | Democratic Party (United States) |
| Home State1: | New Jersey |
| Running Mate1: | George H. Pendleton |
| Electoral Vote1: | 3 |
| Popular Vote1: | 8,767 |
| Percentage1: | 51.81% |
| Nominee2: | Abraham Lincoln |
| Party2: | National Union Party (United States) |
| Home State2: | Illinois |
| Running Mate2: | Andrew Johnson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 8,155 |
| Percentage2: | 48.19% |
| Map Size: | 210px |
| President | |
| Before Election: | Abraham Lincoln |
| Before Party: | Republican Party (United States) |
| After Election: | Abraham Lincoln |
| After Party: | National Union Party (United States) |
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]
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]