10. The graph of the quadratic function $f(x)=a x^{2}+(3 a-1) x-(10 a+3)(a \neq 0)$ always passes through two fixed points. The coordinates of these two points are $\qquad$ .
10. $(2,-5)$ and $(-5,2)$.
$y=\left(x^{2}+3 x-10\right) a-(x+y+3)$, if the fixed point is independent of $a$, then $\left\{\begin{array}{l}x^{2}+3 x-10=0 \\ x+y+3=0,\end{array}\right.$ we get $\left\{\begin{array}{l}x=2 \\ y=-5\end{array}\right.$ or $\left\{\begin{array}{l}x=-5 \\ y=2\end{array}\right.$