Решить неравенство: 12x^2+3x>=0.

\[12x^{2} + 3x \geq 0\]

\[12x\left( x + \frac{3}{12} \right) \geq 0\]

\[x = 0;\ \ x = - \frac{1}{4};\]

\[(x + 0,25) \cdot x \geq 0\]

\[x \leq - 0,25;\ \ x \geq 0.\]

\[Ответ:x \leq - 0,25;\ \ x \geq 0.\]