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x2 + x + 20 = 0; a = 1, b = 1, and c = 20; use the quadratic formula

x = [-b ± √(b2 - 4ac)]/(2a)

x = [-1 ± √(12 - 4*1*20)]/(2*1)

x = (-1 ± √-79)]/2

x = (-1 ± i√79)]/2

The solution set: {-1 - i√79, -1 + i√79}.

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11y ago
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Q: What is the solution set of x2 x 20 equals 0?
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