10. Let $p$ be a prime number, and let the equation $x^{2}-p x-580 p=0$ have two integer roots. Then the value of $p$ is $\qquad$ .
10.29.

Since $x^{2}-p x-580 p=0$ has two integer roots, we have
$$
\Delta=(-p)^{2}+4 \times 580 p=p(p+2320)
$$

is a perfect square.
From this, we know that $p 12320$.
Since $2320=2^{4} \times 5 \times 29$, we have $p=2$ or 5 or 29.
When $p=2$, $\Delta=2^{2}(1+1160)=2^{2} \times 1161=2^{2} \times 3^{2} \times 129$ is not a perfect square;

When $p=5$, $\Delta=5^{2}(1+464)=5^{2} \times 465$ is not a perfect square;

When $p=29$, $\Delta=29^{2}(1+80)=29^{2} \times 9^{2}$, the roots of the equation are $x=\frac{29 \pm 29}{2} \times 9$, i.e., $x_{1}=29 \times 5=145, x_{2}=-29 \times 4=-116$, which satisfies the condition.