How do you solve #m^2+ 2m - 24 = 0#?

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Answer 1 · 2016-04-17T17:45:51

The solutions are:

#m = 4#
# m= -6#

#m^2 + 2m - 24 = 0#

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

#a=1, b=2, c= - 24#

The Discriminant is given by:

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

# = (2)^2-(4 * 1 * (-24))#

# = 4 + 96 = 100#

The solutions are found using the formula:

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

#m= ((-2)+-sqrt(100))/(2*1) = ((-2+-10))/2#

#m= ( -2 +10) /2 = 8/2 = 4#

#m= ( -2 - 10) /2 = -12/2 = -6#