1. Given that $a$, $b$, and $c$ are three distinct odd prime numbers, the equation $(b+c) x^{2}+(a+1) \sqrt{5} x+225=0$ has two equal real roots.
(1) Find the minimum value of $a$;
(2) When $a$ reaches its minimum value, solve this equation.
1. (1) From $\Delta=5(a+1)^{2}-900(b+c)=0$, we get $(a+1)^{2}=2^{2} \times 3^{2} \times 5(b+c)$.
Therefore, $5(b+c)$ should be an even perfect square, with the smallest value being $5^{2} \times$ $2^{2},$ and the smallest value of $a+1$ being $60,$ so the smallest value of $a$ is $59$.
(2) When $a=59$, $b+c=20$.
If $b=3, c=17$, then the original equation is
$20 x^{2}+60 \sqrt{5} x+225=0$.

Solving for $x$ gives $x=-\frac{3}{2} \sqrt{5}$.