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Rozwiązanie równania
[tex]\frac{2x+1}{x-8}=x-22; \ x\neq8\\\\(x-8)(x-22)=2x+1\\\\x^2-22x-8x+176=2x+1\\\\x^2-30x+176=2x+1\\\\x^2-30x+176-2x-1=0\\\\x^2-32x+175=0\\\\a=1, \ b=-32, \ c=175\\\\\Delta=b^2-4ac\rightarrow(-32)^2-4\cdot1\cdot175=1024-700=324\\\\\sqrt{\Delta}=\sqrt{324}=18\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\rightarrow\frac{-(-32)-18}{2\cdot1}=\frac{32-18}{2}=\frac{14}{2}=\boxed7\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\rightarrow\frac{-(-32)+18}{2\cdot1}=\frac{32+18}{2}=\frac{50}{2}=\boxed{25}[/tex]
