# Solve for: x\in (-infinity,-2]\cup [6,infinity)

## Solve for x: $x\in (-\infty,-2]\cup [6,\infty)$

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 $ax^{2}+bx+c=0$.

$$x^{2}-4x-12=0$$

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 $a$, $-4$ for $b$, and $-12$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{4±8}{2}$$

Solve the equation $x=\frac{4±8}{2}$ when $±$ is plus and when $±$ is minus.

$$x=6$$ $$x=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-6\right)\left(x+2\right)\geq 0$$

For the product to be $≥0$, $x-6$ and $x+2$ have to be both $≤0$ or both $≥0$. Consider the case when $x-6$ and $x+2$ are both $≤0$.

$$x-6\leq 0$$ $$x+2\leq 0$$

The solution satisfying both inequalities is $x\leq -2$.

$$x\leq -2$$

Consider the case when $x-6$ and $x+2$ are both $≥0$.

$$x+2\geq 0$$ $$x-6\geq 0$$

The solution satisfying both inequalities is $x\geq 6$.

$$x\geq 6$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x\geq 6$$

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