How do you solve #x^2 - 5x - 6 = 0#?

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Answer 1 · 2016-04-07T09:16:30

The solutions are:

#x = 6#
#x = -1#

#x^2 - 5x - 6 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=-5, c=-6#

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

# = (-5)^2-(4* 1 * (-6))#

# = 25 + 24 = 49#

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

#x = (-(-5)+-sqrt(49))/(2*1) = (5+-7)/2#

#x = (5+7) / 2 = 12/2 = 6#

#x = (5-7) / 2 = -2/2 = -1#