How do you solve # x^2 - 2x + 4# using the quadratic formula?

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Answer 1 · 2015-07-05T13:51:18

#x^2-2x+4 = 0# has no real solutions, only complex ones, but the quadratic formula does work to give:

#x = 1+-isqrt(3)#

#x^2-2x+4# is of the form #ax^2+bx+c#

with #a=1#, #b=-2# and #c=4#

The quadratic formula gives us the solutions to #x^2-2x+4# as:

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

#=(2 +-sqrt(2^2-(4xx1xx4)))/2#

#=(2+-sqrt(-12))/2#

#=(2+-2sqrt(-3))/2#

#=1+-isqrt(3)#