Solve x^2-13x+42geq0?

$$x ^ { 2 } - 13 x + 42 \geq 0 ?$$

Solve for x

$$x\in (-\infty,6]\cup [7,\infty)$$

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.

$$x^{2}-13x+42=0$$

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, -13 for b, and 42 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 1\times 42}}{2}$$

Do the calculations.

$$x=\frac{13±1}{2}$$

Solve the equation x=\frac{13±1}{2} when ± is plus and when ± is minus.

$$x=7 x=6$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-7\right)\left(x-6\right)\geq 0$$

For the product to be ≥0, x-7 and x-6 have to be both ≤0 or both ≥0. Consider the case when x-7 and x-6 are both ≤0.

$$x-7\leq 0 x-6\leq 0$$

The solution satisfying both inequalities is x\leq 6.

$$x\leq 6$$

Consider the case when x-7 and x-6 are both ≥0.

$$x-6\geq 0 x-7\geq 0$$

The solution satisfying both inequalities is x\geq 7.

$$x\geq 7$$

The final solution is the union of the obtained solutions.

$$x\leq 6\text{; }x\geq 7$$