[tex]\frac{x}{x-4} = x-3 \ \ /\cdot(x-4)\\\\\\Zal:\\x-4 \neq 0, \ \ \Rightarrow \ \ x \neq 4\\\\\\(x-3)(x-4) =x\\\\x^{2}-4x-3x+12 = x\\\\x^{2}-7x-x+12 = 0\\\\\underline{x^{2}-8x+12 = 0}\\\\a = 1, \ \ b = -8, \ \ c = 12\\\\\Delta = b^{2}-4ac = (-8)^{2}-4\cdot1\cdot12 = 64 -48 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-8)-4}{2\cdot1} = \frac{4}{2}=2\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-8)+4}{2} = \frac{12}{2} = 6\\\\\underline{x \in \{2, 6\}}[/tex]
