\(x^2-3x+1=0\)
\(a=1,b=-3,c=1\)
\(\Delta=b^2-4ac\)\(=\left(-3\right)^2-4\times1\times1\)\(=5>0\)
\(\Rightarrow pt\)CÓ 2 NO PHÂN BIỆT
\(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{3-\sqrt{5}}{2}\)\(;x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{3+\sqrt{5}}{2}\)
VẬY....