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Решите задачу:

$\frac{x-7}{x-5}+\frac{20}{x^{2}-5^{2}} - \frac{6}{x+5} = 0 \\ \frac{x-7}{x-5}+\frac{20}{(x-5)(x+5)} - \frac{6}{x+5} = 0 \\ \frac{(x-7)(x+5)+20-6(x-5)}{(x-5)(x+5)} = 0 \\ \frac{x^{2}-2x-35+20-6x+30}{(x-5)(x+5)} = 0 \\ \left \{ {{x^{2}-8x+15=0}} \atop {(x-5)(x+5)}} \right. \\ x^{2}-8x+15=0 \\ D = b^{2}-4ac = 8^{2} - 4*1*15 = 64 - 60 = \sqrt{4} = 2 \\ x_{1/2} = \frac{8+/-2}{2} \\ x_{1} = \frac{10}{2} = 5 \\ x_{2} = \frac{6}{2} = 3$