1936 United States presidential election in Delaware

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

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.

  1. ^ "1936 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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