Решите уравнение: (3x^2-2x)/4-(2x^2-2)/3=0.

\[\frac{3x^{2} - 2x}{4} - \frac{2x^{2} - 2}{3} = 0\ \ \ \ \ \ \ | \cdot 12\]

\[3 \bullet \left( 3x^{2} - 2x \right) - 4 \bullet \left( 2x^{2} - 2 \right) =\]

\[= 0\]

\[9x^{2} - 6x - 8x^{2} + 8 = 0\ \]

\[x^{2} - 6x + 8 = 0\]

\[D = ( - 6)^{2} - 4 \cdot 1 \cdot 8 =\]

\[= 36 - 32 = 4\]

\[x_{1} = \frac{6 + \sqrt{4}}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4\]

\[x_{2} = \frac{6 - \sqrt{4}}{2} = \frac{6 - 2}{2} = \frac{4}{2} = 2\]

\[Ответ:x = 4;x = 2.\]