How do you solve #x^2 = 8x – 12#?

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Answer 1 · 2015-06-07T09:02:07

#x²=8x-12#
#x²-8x+12=0# ( we substract #8x-12# on each side )
This is an equation of the form #ax²+bx+c=0#
#a=1#
#b=-8#
#c=12#

We can know calculate #Delta# :

#Delta=b²-4ac#
#Delta=(-8)²-4*1*12#
#Delta=64-48#
#Delta=16# (#=4²#)

#x_1=(-b-sqrtDelta)/(2a)#

#x_1=(8-4)/2#

#x_1=4/2#

#x_1=2#

#x_2=(-b+sqrtDelta)/(2a)#

#x_2=(8+4)/2#

#x_2=12/2#

#x_2=6#

Thus : #x=2# or #x=6#