Solve x^2-5x+6geq0

$$x ^ { 2 } - 5 x + 6 \geq 0$$

Solve for x

$$x\in (-\infty,2]\cup [3,\infty)$$

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.

$$x^{2}-5x+6=0$$

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 1 for a, -5 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{5±1}{2}$$

Solve the equation x=\frac{5±1}{2} when ± is plus and when ± is minus.

$$x=3 x=2$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be ≥0, x-3 and x-2 have to be both ≤0 or both ≥0. Consider the case when x-3 and x-2 are both ≤0.

$$x-3\leq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\leq 2.

$$x\leq 2$$

Consider the case when x-3 and x-2 are both ≥0.

$$x-2\geq 0 x-3\geq 0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq 2\text{; }x\geq 3$$