It doesn't, actually.
Expang the right hand side to get
[tex](x-1)(x-2)=x^2-3x+2[/tex]
So, the equation becomes
[tex]2x+4=x^2-3x+2[/tex]
Move everything to one side to get
[tex]x^2-5x-2=0[/tex]
This equation has solutions
[tex]x_{1,2}=\dfrac{5\pm\sqrt{33}}{2}[/tex]
Which are actually two different real roots.
