Solve 2x^2+3xgeq0

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

Solve for x

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

Solution Steps

Factor out x.

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

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

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

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

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

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

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

The solution satisfying both inequalities is x\geq 0.

$$x\geq 0$$

The final solution is the union of the obtained solutions.

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