#### Answer

$\text{$x-$intercepts: } \{
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^2-2x+3
,$ is/are
\begin{array}{l}\require{cancel}
-x^2-2x+3=0
\\\\
x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(-1)(3)}}{2(-1)}
\\\\
x=\dfrac{2\pm\sqrt{4+12}}{-2}
\\\\
x=\dfrac{2\pm\sqrt{16}}{-2}
\\\\
x=\dfrac{2\pm4}{-2}
.\end{array}
Hence, the $x-$intercepts are
\begin{array}{l}\require{cancel}
\text{$x-$intercepts: } \{
1,-3 \}
.\end{array}