Solve -2x^2+4xleq0

$$-2 { x }^{ 2 } +4x \leq 0$$

Solve for x

$$x\in (-\infty,0]\cup [2,\infty)$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}+4x positive. Since -1 is negative, the inequality direction is changed.

$$2x^{2}-4x\geq 0$$

Factor out x.

$$2x\left(x-2\right)\geq 0$$

For the product to be ≥0, x and x-2 have to be both ≤0 or both ≥0. Consider the case when x and x-2 are both ≤0.

$$x\leq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

Consider the case when x and x-2 are both ≥0.

$$x-2\geq 0 x\geq 0$$

The solution satisfying both inequalities is x\geq 2.

$$x\geq 2$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq 2$$