What is the x in (4/x+2)+(1/x-2) = 5x-6/x^2-4?

$${\frac{{\mathtt{5}}}{{\mathtt{x}}}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{6}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{\mathtt{91}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{225}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{15}}}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{91}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{225}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{15}}}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{91}}}{\left({\mathtt{225}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{8\,434}}}}}{\left({\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2\,539}}}{{\mathtt{3\,375}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{15}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{0.478\: \!747\: \!068\: \!016\: \!753\: \!2}}{\mathtt{\,-\,}}{\mathtt{0.673\: \!492\: \!074\: \!363\: \!498\: \!9}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{0.478\: \!747\: \!068\: \!016\: \!753\: \!2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.673\: \!492\: \!074\: \!363\: \!498\: \!9}}{i}\\
{\mathtt{x}} = {\mathtt{1.757\: \!494\: \!136\: \!033\: \!506\: \!4}}\\
\end{array} \right\}$$

Yuk!  Let's look at this

$${\frac{{\mathtt{5}}}{{\mathtt{x}}}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{6}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}$$

There are no real roots.

More inspection here.

http://www.wolframalpha.com/input/?i=5%2Fx%3D5x-6%2Fx%5E2-4

That is the best I can do I think. 

after you clear the denominator (with (x+2)(x-2)), you get 4x-8+x+2=5x-6 then you add to get everything that cancels, so therefore 0 must equal 0.

Is this what you mean or do you need to add some brackets.

$$\left({\frac{{\mathtt{4}}}{{\mathtt{x}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{x}}}}{\mathtt{\,-\,}}{\mathtt{2}}\right) = {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{6}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}$$

$${\frac{{\mathtt{5}}}{{\mathtt{x}}}} = {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{6}}}{{{\mathtt{x}}}^{{\mathtt{2}}}}}{\mathtt{\,-\,}}{\mathtt{4}}$$

multiply everything by x^2

i wont do any more unless you request it because this is likely not the intended question. 

Yes, @Melody. That's right. :) PLEASE ANSWER