How do you solve #x^2-2x-12=0# algebraically?

The whole number factors of 12 are {6,2}, {3,4} and none of these have a difference of 2 ( from #2x#). So it must have none integer solutions for #x#. Thus we need to use the formulas. These formula are something worth remembering.

It takes quite a bit of work to make them stick!

Standard equation form:#" "y=ax^2 +bx+c#

where #color(magenta)(x=(-b+-sqrt(b^2-4ac))/(2a))# ............................(1)

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Another one worth remembering is the completed square format.
Otherwise known as Vertex Form.

#" "color(magenta)(y=a(x+b/(2a))^2+c -[(b/2)^2])# .............................(2)

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#color(blue)("Using equation (1)")#

#=>x" "=" "(-(-2)+-sqrt( (-2)^2-4(1)(-12)))/(2(1))#

#=>x" "=" "(2+-sqrt( 52))/2#

#=>x" "=" "(2+-sqrt(2^2xx13))/2#

#color(green)(=>x= 1+-sqrt(13))#
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#color(blue)("Using equation (2)")#

#y=1(x-1)^2-12-1#

#y=(x-1)^2-13#

Set #y=0#

#0=(x-1)^2-13#

Add 13 to both sides

#13=(x-1)^2#

Square root both sides

#+-sqrt(13)=x-1#

Add 1 to both sides

#color(green)(x=1+-sqrt(13))#
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#x~~4.606" " -2.606# to 3 decimal places