10. The system of equations
$$
\left\{\begin{array}{l}
a+b+c+d=-2, \\
a b+a c+a d+b c+b d+c d=-3, \\
b c d+a c d+a b d+a b c=4, \\
a b c d=3
\end{array}\right.
$$

has a set of real solutions $(a, b, c, d)=$ . $\qquad$
10. $\left(\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}\right)$.

By Vieta's formulas, $a, b, c, d$ are exactly the four real roots of the equation
$$
x^{4}+2 x^{3}-3 x^{2}-4 x+3=0
$$

Notice that,
$$
\begin{array}{l}
x^{4}+2 x^{3}-3 x^{2}-4 x+3=0 \\
\Rightarrow\left(x^{2}+x+1\right)^{2}-6\left(x^{2}+x+1\right)+8=0 . \\
\Rightarrow x^{2}+x+1=2 \text { or } x^{2}+x+1=4 .
\end{array}
$$

Solving the system of equations, the four real solutions are
$$
\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}
$$

in some permutation.