Algebra

$\dfrac{x+4}{x+5} = \dfrac{x-3}{2} + \dfrac{x + 7}{5}\\ \dfrac{10(x+4)}{10(x+5)}=\dfrac{(x+5)(x-3)+(x+5)(x+7)}{10(x+5)}\\ 10x+40=x^2+2x-15+x^2+12x+35\\ 2x^2+4x-20=0\\ x^2-2x-10=0\\ x_{1,2}=1\pm \sqrt{1+10}\\ \color{blue}x\in \{-2.3166, 4.3166\}$

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