Решите уравнение: (x(x-1))/4+(x-3)^2/2=(4-x)^2/3-1/3.

\[3x(x - 1) + 6\left( x^{2} - 6x + 9 \right) =\]

\[= 4 \cdot \left( 16 - 8x + x^{2} \right) - 4\]

\[3x^{2} - 3x + 6x^{2} - 36x + 54 =\]

\[= 64 - 32x + 4x^{2} - 4\]

\[9x^{2} - 4x^{2} - 39x + 32x + 54 - 60 =\]

\[= 0\]

\[5x^{2} - 7x - 6 = 0\]

\[D = 49 + 120 = 169\]

\[x_{1} = \frac{7 + 13}{10} = 2;\ \ \ \]

\[x_{2} = \frac{7 - 13}{10} = - \frac{6}{10} = - 0,6\]

\[Ответ:x = 2;\ \ x = 0,6.\]