How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.

We have
$$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$ and 
$$\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)\left(
\sqrt{49-x^2}+\sqrt{25-x^2}\right)=3\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right),$$
which gives 
$$\sqrt{49-x^2}+\sqrt{25-x^2}=8$$ and from here
$$\sqrt{49-x^2}=5.5,$$ which gives the answer:
$$\left\{-\sqrt{18.75},\sqrt{18.75}\right\}.$$
