Given: #0=x^2-6x+12#
#color(brown)("The question specifically equates to 0. So we must find a solution that")##color(brown)("works for "y=0. #
Consider #y=ax^2+bx+c color(white)("ddd")and color(white)("ddd")x=(-b+-sqrt(b^2-4ac))/(2a)#
Now lets look at the equation part : #b^2-4ac#. THis is called the determinant.
#b^2-4ac color(white)("ddd")->color(white)("d")
color(white)("ddd") (-6)^2-4(1)(12) = -12#
As this is negative the graph does not have any x-intercepts where #x# is in the set of what is called real numbers #x !in RR#
There will be a solution of #x# where it is in the set of 'Complex Numbers. #x in CC#
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#x=(-b+-sqrt(b^2-4ac))/(2a) color(white)("d")->color(white)("d")x=(+6+-sqrt(-12))/2#
#x=3+-sqrt((-12)/4)#
#x=3+-sqrt(3xx(-1))#
#x=3+-sqrt(3)xxsqrt(-1)#
But #sqrt(-1)=i#
#x=3+-sqrt(3)color(white)(..)i#
