Solve (2)3x^2+4xgeq0

$$( 2 ) 3 x ^ { 2 } + 4 x \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{2}{3}]\cup [0,\infty)$$

Solution Steps

Multiply 2 and 3 to get 6.

$$6x^{2}+4x\geq 0$$

Factor out x.

$$2x\left(3x+2\right)\geq 0$$

For the product to be ≥0, x+\frac{2}{3} and x have to be both ≤0 or both ≥0. Consider the case when x+\frac{2}{3} and x are both ≤0.

$$x+\frac{2}{3}\leq 0 x\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{2}{3}.

$$x\leq -\frac{2}{3}$$

Consider the case when x+\frac{2}{3} and x are both ≥0.

$$x\geq 0 x+\frac{2}{3}\geq 0$$

The solution satisfying both inequalities is x\geq 0.

$$x\geq 0$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{2}{3}\text{; }x\geq 0$$