1884 United States presidential election in Delaware
Election in Delaware
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 County Results Cleveland 50-60% 60-70%
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The 1884 United States presidential election in Delaware took place on November 4, 1884, as part of the 1884 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, Grover Cleveland, over the Republican nominee, James G. Blaine. Cleveland won the state by a margin of 13.35%.
Results
| 1884 United States presidential election in Delaware[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Grover Cleveland of New York | Thomas Andrews Hendricks of Indiana | 16,957 | 56.55% | 3 | 100.00% | ||
| Republican | James Gillespie Blaine of Maine | John Alexander Logan of Illinois | 12,953 | 43.20% | 0 | 0.00% | ||
| Prohibition | John Pierce St. John of Kansas | William Daniel of Maryland | 64 | 0.21% | 0 | 0.00% | ||
| Greenback | Benjamin Franklin Butler of Massachusetts | Absolom Madden West of Mississippi | 10 | 0.03% | 0 | 0.00% | ||
| Total | 29,984 | 100.00% | 3 | 100.00% | ||||
See also
References
- ^ "1884 Presidential General Election Results - Delaware". U.S. Election Atlas. Retrieved December 23, 2013.
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