#### Answer

$\{ 1, 3 \}$

#### Work Step by Step

Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $
y=x^2-4x+3
,$ is/are
\begin{array}{l}\require{cancel}
x^2-4x+3=0
\\\\
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}
\\\\
x=\dfrac{4\pm\sqrt{16-12}}{2}
\\\\
x=\dfrac{4\pm\sqrt{4}}{2}
\\\\
x=\dfrac{4\pm2}{2}
.\end{array}
Hence, the $x-$intercepts are $
\{ 1, 3 \}
.$