algebra precalculus - Solve $sqrt{3x}+sqrt{2x}=17$

This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:

$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.