1

We have

$$2025x^2+ax-3=0$$

and

$$\Delta =a^2-4\times 2025\times (-3)=a^2+24300$$

We know that $\Delta$ is a perfect square, then there’s an integer $k$ which

$$a^2+24300=k^2⟹k^2-a^2=24300⟹(k-a)(k+a)=24300$$

We know that

$$24300=2^2\times 3^5\times 5^2$$

We should choose 2 factors in which there difference is the smallest. We choose

$$b_1=2\times 3\times 5^2=150,b_2=2\times 3^4=162$$ $$150\times 162=24300.$$

So we have

$$k=\frac{150+162}{2}=156,a=\frac{162-150}{2}=6$$

Hence, the answer is

$$\boxed{6}$$