Решите уравнение методом замены переменной: 6/(x^2-3x+5)-x^2+3x=4.

\[\frac{6}{x^{2} - 3x + 5} - x^{2} + 3x = 4\]

\[Пусть\ \ t = x^{2} - 3x + 5:\ \]

\[\frac{6}{t} - t + 1 = 0\ \ \ \ \ \ \ | \cdot t\]

\[- t^{2} + t + 6 = 0\]

\[t^{2} - t - 6 = 0\]

\[t_{1} + t_{2} = 1;\ \ t_{1} \cdot t_{2} = - 6\]

\[t_{1} = 3,\ \ t_{2} = - 2.\]