1936 United States presidential election in Delaware information
Election in Delaware
Main article: 1936 United States presidential election
1936 United States presidential election in Delaware
← 1932
November 3, 1936
1940 →
Nominee
Franklin D. Roosevelt
Alf Landon
Party
Democratic
Republican
Home state
New York
Kansas
Running mate
John Nance Garner
Frank Knox
Electoral vote
3
0
Popular vote
69,702
57,236
Percentage
54.62%
44.85%
County Results
Roosevelt
50-60%
President before election
Franklin D. Roosevelt
Democratic
Elected President
Franklin D. Roosevelt
Democratic
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v
t
e
The 1936 United States presidential election in Delaware was held on November 3, 1936. The state voters chose three electors to the Electoral College, who voted for president and vice president.
Delaware voted for Democratic Party candidate and incumbent President Franklin D. Roosevelt, who defeated Republican nominee, Kansas Governor Alf Landon. Roosevelt won the state by a margin of 9.77%, marking the first time since 1912 that the state voted for a Democratic presidential candidate, and the first time since 1888 that a Democrat carried the state with an outright majority.
While Landon lost the state, the 44.85% of the popular vote made Delaware his fifth strongest state in the 1936 election in terms of popular vote percentage after Vermont, Maine, New Hampshire and Kansas.[1]
This election marked Delaware's transition into a bellwether state: for the rest of the 20th century, it would vote for a losing candidate only once (in 1948). Once the century rolled around, it came to be regarded as a solidly blue state.
^"1936 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
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