How do you solve #x^2 +2x -4=0#?

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Answer 1 · 2015-09-12T07:14:38

The solutions are
#color(blue)(x=-1+sqrt(5)#
#color(blue)(x=-1-sqrt(5)#

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

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=2, c=-4#

The Discriminant is given by:
#Delta=b^2-4*a*c#

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

# = 4 +16 = 20#

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

#x = ((-2)+-sqrt(20))/(2*1) = (-2+-2sqrt(5))/2#

#=(cancel2(-1+-sqrt(5)))/cancel2#

The solutions are
#color(blue)(x=-1+sqrt(5)#
#color(blue)(x=-1-sqrt(5)#