Algebra

Find all real x.

$\color{blue}2 \cdot \frac{x - 5}{x - 3} > \frac{2x - 5}{x + 2} + 20\\ f(x)=2 \cdot \frac{x - 5}{x - 3} - \frac{2x - 5}{x + 2} -20>0\\ 2(x-5)(x+2)-(2x-5)(x-3)-20(x-3)(x+2)>0\\ (2x-10)(x+2)-(2x^2-6x-5x+15)-20(x^2+2x-3x-6)>0$

$2x^2+4x-10x-20-2x^2+11x-15-20x^2-40x+60x+120>0\\ f(x)=-20x^2+25x+85>0$

$x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} = {-25 \pm \sqrt{25^2+4\cdot 20\cdot 85} \over -40}\\ x_0\in \{-1.5292,2.7792\}\\$

$In\ the\ calculated\ range\ is\\ (2 \cdot \frac{x - 5}{x - 3} ){\color{red}\ <}\ (\frac{2x - 5}{x + 2} + 20)\\ \color{blue}x\in \mathbb R\ (-1.52921099245< x< 2.77921099245)$

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