\(\left \{ {{y=x^2-4x-5} \atop {y+x=-1}} \right. \\
\\
\left \{ {{y=x^2-4x-5} \atop {y=-1-x}} \right. \\\\
\\
x^2-4x-5=-1-x\\
\\
x^2-3x-4=0\\
\\
\Delta=(-3)^2-4.1.(-4)=9+15=25\\
\\
x=\frac{3 \pm5}{2}\\
\\
x_1=-1 \rightarrow y=-1+1=0\\
\\
x_2=4 \rightarrow y=-1-4=-5\)
We found 2 points that satisfies both equations (4,-5) and (-1,0)
