Решите уравнение: (x^2+3x)/2+(x-3x^2)/8=2x.

\[\frac{x^{2} + 3x}{2} + \frac{x - 3x^{2}}{8} = 2x\ \ \ \ | \cdot 8\]

\[4 \cdot \left( x^{2} + 3x \right) + x - 3x^{2} = 16x\]

\[4x^{2} + 12x + x - 3x^{2} - 16x = 0\]

\[x² - 3x = 0\]

\(x(x - 3) = 0\)

\[x = 0\ \ \ \ \ \ \ \ x - 3 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 3\]

\[Ответ:\ \ x = 0\ \ \ \ и\ \ \ x = 3.\]