How do you solve #x^2-2x+5=0# using the quadratic formula?

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Answer 1 · 2016-02-23T17:43:29

The solutions are:
# x= (2+4i)/2#

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

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

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=1, b=-2, c=5#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-2)^2-(4*1*5)#

# = 4 -20#

#=-16#

The solutions are found using the formula:
#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-2)+-sqrt(-16))/(2*1) = (2+-sqrt(-16))/2#

# = (2+-sqrt(-1 xx 16))/2#

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

The solutions are:
# x= (2+4i)/2#

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