You mean solve? To solve the equation, first of all note that #2/1=2#, and that we will not accept #x=-12# as a solution, because it would annihilate the denominator.
So, if #x# is not #-12#, we can multiply the whole expression by #x+12#, obtaining
#x^2=2(x+12)#. Expanding the right member we get
#x^2=2x+24#, and bringing everything to the left member we have
#x^2-2x-24=0#, which is a classical quadratic equation of the form
#ax^2+bx+c=0#, where #a=1#, #b=-2# and #c=-24#.
The discriminant of the quadratic is #b^2-4ac#, in your case 4-4*(-24)= 4+96=100#, which is positive, so the equation has two solutions, namely
#x_{1,2} = \frac{-b\pm\sqrt(b^2-4ac)}{2a}=
(2pmsqrt(100))/2=(2pm10)/2=(1pm5)#
So, the two solutions are #-4# and #6#.
