How do you solve #x^2=6x-12#?

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Answer 1 · 2016-06-28T14:08:57

#x=3+-sqrt3 i#

let's rewrite the equation in the general form:

#x^2-6x+12=0#

and solve by the quadratic formula used when b is even and a=1

#x=(-b/2+-sqrt((b/2)^2-c))#

In this case a=1; b=-6; c=12, so

#x=3+-sqrt(9-12)#

#x=3+-sqrt3 i#

no real solution, but complex