$$x^{2}-4x-12=0$$
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.$$x\leq -2\text{; }x\geq 6$$