$$ x^2-4x+3=0$$
\(
\begin{array}{ccc} & & x_1 = \frac{4+2}{2}=3\\ & \nearrow &\\ x=\frac{-(-4)\pm \sqrt{(-4)^2-4 \cdot1\cdot3}}{2 \cdot1}=
\frac{4\pm \sqrt{4}}{2}& &\\ & \searrow &\\& &x_2 = \frac{4-2}{2}=1\end{array}
\)
Las soluciones son \(x=3\quad \) y \(\quad x=1\)