Помогите решите уравнение.

$\frac{\textup{x}}{\textup{x}-1} + \frac{\textup{x}}{\textup{x}+3} = \frac{1}{(\textup{x}-1)(\textup{x}+3)} \\ \\ \frac{\textup{x}(\textup{x}+3)+\textup{x}(\textup{x}-1)}{(\textup{x}-1)(\textup{x}+3)} = \frac{1}{(\textup{x}-1)(\textup{x}+3)} ~~|\cdot(\textup{x}-1)(\textup{x}+3) \\ \\ \textup{x}(\textup{x}+3)+\textup{x}(\textup{x}-1)=1 \\ \textup{x}^2+3\textup{x}+\textup{x}^2-1-1=0 \\ 2\textup{x}^2+3\textup{x}-2=0 \\ \textup{D}=9-4\cdot2\cdot(-2)=9+16=25$

$\textup{x}_1= \frac{-3+5}{4} = \frac{2}{4} =0,5 \\ \\ \textup{x}_2= \frac{-3-5}{4} = \frac{-8}{4} =-2$

Ответ: $-2;~0,5$