4. If $x_{1}$ and $x_{2}$ are the two distinct real roots of the equation $\sqrt{x+4}+\sqrt{9-3 x}=5$, then, the value of $x_{1}^{2}+x_{2}^{2}$ is $\qquad$ .
4. $14 \frac{1}{16}$.

The original equation, when squared and simplified on both sides, yields
$$
\sqrt{x+4} \cdot \sqrt{9-3 x}=x+6 \text {, }
$$

Squaring both sides again gives
$$
9 x+36-3 x^{2}-12 x=x^{2}+12 x+36 \text {, }
$$

Combining terms results in $4 x^{2}+15 x=0$, or $x(4 x+15)=0$.
Solving yields $x_{1}=0, x_{2}=-\frac{15}{4}$.
Verification shows that both are roots of the original equation. Therefore,
$$
x_{1}^{2}+x_{2}^{2}=0+\frac{225}{16}=14 \frac{1}{16} \text {. }
$$