Answer
$\text{$x-$intercepts: } \{
-1-\sqrt{5},-1+\sqrt{5} \}
$
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-4
,$ is/are
\begin{array}{l}\require{cancel}
x^2+2x-4=0
\\\\
x=\dfrac{-2\pm\sqrt{2^2-4(1)(-4)}}{2(1)}
\\\\
x=\dfrac{-2\pm\sqrt{4+16}}{2}
\\\\
x=\dfrac{-2\pm\sqrt{20}}{2}
\\\\
x=\dfrac{-2\pm\sqrt{4\cdot5}}{2}
\\\\
x=\dfrac{-2\pm\sqrt{(2)^2\cdot5}}{2}
\\\\
x=\dfrac{-2\pm2\sqrt{5}}{2}
\\\\
x=\dfrac{2(-1\pm\sqrt{5})}{2}
\\\\
x=\dfrac{\cancel{2}(-1\pm\sqrt{5})}{\cancel{2}}
\\\\
x=-1\pm\sqrt{5}
.\end{array}
Hence, the $x-$intercepts are
\begin{array}{l}\require{cancel}
\text{$x-$intercepts: } \{
-1-\sqrt{5},-1+\sqrt{5} \}
.\end{array}
