Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square of both sides and I tried to factor and failed.

The idea is to move the equation's terms around such that one and only one square root appears on one side. Squaring then removes that isolated square root; the other side may still have a square root, but there will be one less overall. Repeat until a polynomial is obtained.
Let $y=x^2$, then (using fractions instead of decimals) we have
$$\sqrt{\frac{49}4+y}-\sqrt{9+y}=\frac14$$
$$\sqrt{\frac{49}4+y}=\sqrt{9+y}+\frac14$$
$$\frac{49}4+y=\left(\sqrt{9+y}+\frac14\right)^2=9+y+\frac12\sqrt{9+y}+\frac1{16}$$
$$\frac{49}4-9-\frac1{16}=\frac12\sqrt{9+y}=\frac{51}{16}$$
$$\sqrt{9+y}=\frac{51}{8}$$
$$9+y=\frac{2601}{64}$$
$$y=\frac{2025}{64}$$
$$x=\pm\frac{45}8=\pm5.625$$
