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

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Answer 1 · 2016-04-24T07:35:32

The solutions are:
# x = 4#

# x = -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 = 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 = 8/2 = 4#

# x = (2- 6)/2 = (-4)/2 = -2#