How do you solve #x^2+2x+8=0#?

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Answer 1 · 2016-02-26T15:53:51

The solutions are:
#x= (-2+sqrt(-28))/2#

#x=(-2-sqrt(-28))/2#

#x^2+2x+8=0#

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

#a=1, b=2, c=8#

The Discriminant is given by:

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

# = (2)^2-(4*1*8)#

# = 4-32=-28#

The solutions are found using the formula:

#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-2)+-sqrt(-28))/(2*1) = (-2+-sqrt(-28))/2#

The solutions are:
#x= (-2+sqrt(-28))/2#

#x=(-2-sqrt(-28))/2#