[tex]\frac{2x-3}{x+1} = x-\frac{5}{3} \ \ |\cdot3\\\\\frac{3(2x-3)}{x+1} = 3x-5 \ \ |\cdot(x+1)\\\\\\Zal:\\x+1 \neq 0 \ \ \rightarrow \ \ x \neq -1\\D = R \setminus\{-1\}\\\\\\(3x-5)(x+1) = 3(2x-3)\\\\3x^{2}+3x-5x-5 = 6x-9\\\\3x^{2}-2x-5 = 6x-9\\\\3x^{2}-2x-6x-5+9 = 0\\\\\underline{3x^{2}-8x+4 = 0}\\\\a = 3, \ b = -8, \ c = 4[/tex]
[tex]\Delta = b^{2}-4ac = (-8)^{2}-4\cdot3\cdot4 = 64 - 48 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{8-4}{2\cdot3} = \frac{4}{6} = \boxed{\frac{2}{3}}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{8+4}{6} = \frac{12}{6} = \boxed{2}\\\\x \in \{\frac{2}{3}, 2\}[/tex]