How do you solve the quadratic with complex numbers given #-x^2+4x-5=0#?

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Answer 1 · 2016-09-10T04:35:27

#-x^2+4x-5=0#

Use the quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=-1#
#b=4#
#c=-5#

#x=(-4+-sqrt(4^2-(4*-1*-5)))/(2*-1)#

#x=(-4+-sqrt(16-20))/-2#

#x=(-4+-sqrt(-4))/-2#

#x=(-4+-sqrt(4)sqrt(-1))/-2#

#sqrt(-1) = i#

#x=(-4+-2i)/-2#

Divide both of the terms in the numerator by the denominator.

#x=(-4)/-2 +- (2i)/-2#

#x=2+-i# which can also be written as #x=2+i# and #x=2-i#