Решите уравнение: (2x^2-3x)/3-(3x^2-12)/5=0.

\[\frac{2x^{2} - 3x}{3} - \frac{3x^{2} - 12}{5} = 0\ \ \ \ \ \ | \cdot 15\]

\[5 \bullet \left( 2x^{2} - 3x \right) - 3 \bullet \left( 3x^{2} - 12 \right) =\]

\[= 0\]

\[10x^{2} - 15x - 9x^{2} + 36 = 0\]

\[x^{2} - 15x + 36 = 0\]

\[D = ( - 15)^{2} - 4 \cdot 1 \cdot 36 =\]

\[= 225 - 144 = 81\]

\[x_{1} = \frac{15 + \sqrt{81}}{2} = \frac{15 + 9}{2} = \frac{24}{2} =\]

\[= 12\]

\[x_{2} = \frac{15 - \sqrt{81}}{2} = \frac{15 - 9}{2} = \frac{6}{2} =\]

\[= 3\]

\[Ответ:12;3.\]