Задача:Решить рациональное уравнение

$\frac{2x^{2}-3x }{x-2}=\frac{3-2x}{2-x}\\\\\frac{2x^{2}-3x }{x-2}-\frac{3-2x}{2-x}=0\\\\\frac{2x^{2}-3x }{x-2}+\frac{3-2x}{x-2}=0\\\\\frac{2x^{2}-3x+3-2x }{x-2}=0\\\\\ \left \{ {{2x^{2}-5x+3=0 } \atop {x-2\neq0 }} \right.\\\\\left \{ {{2x^{2}-5x+3=0 } \atop {x\neq 2}} \right.\\\\2x^{2}-5x+3=0\\\\D=(-5)^{2} -4*2*3=25-24=1\\\\x_{1} =\frac{5+1}{4}=1,5\\\\x_{2}=\frac{5-1}{4}=1$

Ответ : 1,5 ; 1