my question

3/(x-2)-1/(x+2)=5

\(\begin{array}{|rcll|} \hline 5 &=& \frac{3}{x-2}-\frac{1}{x+2} \quad & | \quad \cdot (x+2)(x-2) \\ 5\cdot(x+2)(x-2) &=& \frac{ 3(x+2)(x-2)}{x-2}-\frac{(x+2)(x-2) }{x+2} \\ 5\cdot (x+2)(x-2) &=& 3(x+2) - (x-2) \\ 5\cdot (x+2)(x-2)) &=& 3x+ 6 - x+2 \\ 5\cdot (x+2)(x-2) &=& 2x+8 \\ 5\cdot (x^2-4) &=& 2x+8 \\ 5x^2-20 &=& 2x+8 \quad & | \quad -2x-8 \\ 5x^2-20 -2x-8 &=& 0 \\ 5x^2-2x -28 &=& 0 \\ \\ \begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{-b\pm\sqrt{b^2-4ac} }{2a} \\ \hline \end{array} \\ ax^2+bx+c &=& 0 \quad & | \quad a = 5 \quad b = -2 \quad c=-28 \\ x &=& \frac{ 2\pm\sqrt{4-4\cdot 5 (-28)} }{2\cdot 5} \\ x &=& \frac{2\pm\sqrt{4+20\cdot28)} }{10} \\ x &=& \frac{2\pm\sqrt{564} }{10} \\ x &=& \frac{2\pm 23.7486841741 }{10} \\\\ x_1 & = & \frac{2+ 23.7486841741 }{10} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2.57486841741 } \\\\ x_2 &=& \frac{2- 23.7486841741 }{10} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-2.17486841741} \\ \hline \end{array} \)