Solve 2x^2-5x+2geq0

$$2 { x }^{ 2 } -5x+2 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{1}{2}]\cup [2,\infty)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2x^{2}-5x+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -5 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Do the calculations.

$$x=\frac{5±3}{4}$$

Solve the equation x=\frac{5±3}{4} when ± is plus and when ± is minus.

$$x=2 x=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-2\right)\left(x-\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, x-2 and x-\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when x-2 and x-\frac{1}{2} are both ≤0.

$$x-2\leq 0 x-\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{1}{2}.

$$x\leq \frac{1}{2}$$

Consider the case when x-2 and x-\frac{1}{2} are both ≥0.

$$x-\frac{1}{2}\geq 0 x-2\geq 0$$

The solution satisfying both inequalities is x\geq 2.

$$x\geq 2$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{1}{2}\text{; }x\geq 2$$