<s>[INST] Algebra QF $f(x)=x^2-18x-4$ I have the problem to find the zero of the function:
$$f(x)=x^2-18x-4$$
I have it mostly worked out as $a=1, b= -18, c= -4$
worked out I have:
$$\frac{18\pm \sqrt{340}}{2}$$
I know the answers are 
$$9+ \sqrt{85}\\
9- \sqrt{85}$$
The question is when working the problem out I end up with:
$$\frac{18\pm \sqrt{340}}{2}$$
I am not sure how that step (with  √ 340) becomes $9\pm\sqrt{85}$
Any help would be awesome!
 [/INST] Note that:
$$\frac{18\pm\sqrt{340}}{2}=\frac{18\pm\sqrt{4\cdot85}}{2}=\frac{18\pm\sqrt{4}\cdot\sqrt{85}}{2}=\frac{18\pm2\sqrt{85}}{2}=\frac{2\cdot(9\pm\sqrt{85})}{2}=9\pm\sqrt{85}$$
</s>