What are the intersection points for these two equations? For the sake of me, I can't get the solution! Not sure what I have done wrong. Anyway these are the two equations:
$$\sqrt{25-(x-13.1)^2}=\frac x2-4$$
and I went
$$-(x-13.1)^2=\frac{x^2}4+16-25$$
$$(x-13.1)^2=-\frac{x^2}4+9$$
$$x^2+26.2x-171.61=-\frac{x^2}4-9$$
What have I done wrong? Thank you in advance!!

You've made a (common) mistake on your first line. When expanding $(a+b)^2$ some students only write $a^2+b^2$ where in fact the correct expansion is $a^2+2ab+b^2$.
So you should have gotten $\frac{x^2}{4}-4x+16$ when you expanded $\left(\frac{x}{2}-4\right)^2$ instead of just $\frac{x^2}{4}+16$.
Also on your third line you have your pluses and minuses incorrect. $(x-13.1)^2$ should expand to give $x^2-26.2x+171.61$.
Lastly on your last line you have change $+9$ into $-9$ which I assume is a transcription error.
The correct working out would look something like:
$$-(x-13.1)^2=\frac{x^2}{4}-4x+16-25$$
$$(x-13.1)^2=-\frac{x^2}{4}\color{red}{+4x}+9$$
$$x^2\color{red}{-}26.2x\color{red}{+}171.61=-\frac{x^2}{4}-4x\color{red}{+}9$$
$$\frac{5x^2}{4}+30.2x+162.61=0$$
$$x^2-24.16x+130.088=0$$
$$x=\frac{24.16\pm\sqrt{24.16^2-4\times1\times130.088}}{2}$$
$$x\approx8.10025\text{ or }x\approx16.0597$$
