How do you solve #-x^2 + 5x - 9 = 0#?

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How do you solve #-x^2 + 5x - 9 = 0#?

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Answer 1 · 2016-08-18T02:32:18

#(5 +- isqrt11)/2#

Explanation:

y = - x^2 + 5x - 9 = 0
Use the improved quadratic formula (Socratic Search):
D = d^2 = b^2 - 4ac = 25 - 36 = - 11.
There are no real roots because D < 0. There are 2 imaginary roots.
#D = 11i^2# --> #d = +- isqrt11#

#x = - b/(2a) +- d/(2a) = -5/-2 +- isqrt11/2 = (5 +- isqrt11)/2#

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How do you solve #-x^2 + 5x - 9 = 0#?

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