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

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Answer 1 · 2016-05-21T11:48:08

The solutions for the equation are:
#color(blue)(x = 2#

#color(blue)(x = -4#

#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 = 36#

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

#x = (-2)+-sqrt(36))/(2*1) = ((-2+-6))/2#

#x = (-2 + 6 ) / 2 = 4/2 = 2# , #color(blue)(x = 2#

#x = (-2 - 6 ) / 2 = -8/2 = -4# , #color(blue)(x = -4#