Solve x^2-12x-72geq0

$${ x }^{ 2 } -12x-72 \geq 0$$

Solve for x

$$x\in (-\infty,6-6\sqrt{3}]\cup [6\sqrt{3}+6,\infty)$$

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.

$$x^{2}-12x-72=0$$

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, -12 for b, and -72 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 1\left(-72\right)}}{2}$$

Do the calculations.

$$x=\frac{12±12\sqrt{3}}{2}$$

Solve the equation x=\frac{12±12\sqrt{3}}{2} when ± is plus and when ± is minus.

$$x=6\sqrt{3}+6 x=6-6\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(6\sqrt{3}+6\right)\right)\left(x-\left(6-6\sqrt{3}\right)\right)\geq 0$$

For the product to be ≥0, x-\left(6\sqrt{3}+6\right) and x-\left(6-6\sqrt{3}\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(6\sqrt{3}+6\right) and x-\left(6-6\sqrt{3}\right) are both ≤0.

$$x-\left(6\sqrt{3}+6\right)\leq 0 x-\left(6-6\sqrt{3}\right)\leq 0$$

The solution satisfying both inequalities is x\leq 6-6\sqrt{3}.

$$x\leq 6-6\sqrt{3}$$

Consider the case when x-\left(6\sqrt{3}+6\right) and x-\left(6-6\sqrt{3}\right) are both ≥0.

$$x-\left(6-6\sqrt{3}\right)\geq 0 x-\left(6\sqrt{3}+6\right)\geq 0$$

The solution satisfying both inequalities is x\geq 6\sqrt{3}+6.

$$x\geq 6\sqrt{3}+6$$

The final solution is the union of the obtained solutions.

$$x\leq 6-6\sqrt{3}\text{; }x\geq 6\sqrt{3}+6$$