#### Answer

Solution set is, $\left\{ \left( -4,3 \right),\left( 0,-1 \right) \right\}$.

#### Work Step by Step

Let us consider the provided system:
$\begin{align}
& {{x}^{2}}-4y=4 \\
& x+y=-1
\end{align}$
So, from the second equation:
$y=-1-x$
Substitute $y=-1-x$ in the first equation and solve for x:
$\begin{align}
& {{x}^{2}}-4\left( -1-x \right)=4 \\
& {{x}^{2}}+4+4x=4 \\
& {{x}^{2}}+4x=0 \\
& x\left( x+4 \right)=0
\end{align}$
It implies that:
$x=0,-4$
Substitute $x=0$ in the second equation and find the corresponding value of y:
$\begin{align}
& 0+y=-1 \\
& y=-1
\end{align}$
Substitute $x=-4$ in the second equation and find the corresponding value of y:
$\begin{align}
& -4+y=-1 \\
& y=3
\end{align}$
Thus, that set is $\left\{ \left( -4,3 \right),\left( 0,-1 \right) \right\}$.