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Rewrite the system $\begin{cases} x^2+y+xy(x^2+y)+xy=\frac{-5}{4}\\ (x^2+y)^2+xy=\frac{-5}{4}\end{cases}$ Now we call $x^2+y=t$ and $xy=u$ , the system becomes $\begin{cases} t+u.t+u=\frac{-5}{4}\\ t^2+u=\frac{-5}{4}\end{cases}$ $\Leftrightarrow\begin{cases} t(t-u-1)=0\\ t^2+u=\frac{-5}{4}\end{cases}$ We get $(t;u)=(0;\frac{-5}{4});(\frac{-1}{2};\frac{-3}{2})$ as the answer. So

$1.$ When $(t;u)=(0;\frac{-5}{4})$ , we have $\begin{cases} x^2+y=0\\xy=\frac{-5}{4}\end{cases}$ $\Leftrightarrow (x;y)=\bigg(\sqrt[3]{\frac{5}{4}};-\bigg(\sqrt[3]{\frac{5}{4}}\bigg)^2\bigg)$ $2.$ When $(t;u)=(\frac{-1}{2};\frac{-3}{2})$ , we have $\begin{cases} x^2+y=\frac{-1}{2}\\xy=\frac{-3}{2}\end{cases}$ $\Leftrightarrow (x;y)=\bigg(1;\frac{-3}{2}\bigg)$ So the sum of all x and y will be $1+\sqrt[3]{\frac{5}{4}}-\frac{3}{2}-\bigg(\sqrt[3]{\frac{5}{4}}\bigg)^2\approx -0.58$