Respuesta :
[tex]x^2-3x+2=0\\\\a=1;\ b=-3;\ c=2\\\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot1\cdot2=9-8=1\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt1=1\\\\x_1=\frac{3-1}{2\cdot1}=\frac{2}{2}=1;\ x_2=\frac{3+1}{2\cdot1}=\frac{4}{2}=2\\\\Answer:x=1\ or\ x=2.[/tex]
[tex]x^2-3x+2=(x-1)(x-2)[/tex]
[tex]x^2-3x+2=(x-1)(x-2)[/tex]
