Solve x^2-3x-18>0

$${ x }^{ 2 } -3x-18 > 0$$

Solve for x

$$x\in \left(-\infty,-3\right)\cup \left(6,\infty\right)$$

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.

$$x^{2}-3x-18=0$$

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, -3 for b, and -18 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-18\right)}}{2}$$

Do the calculations.

$$x=\frac{3±9}{2}$$

Solve the equation x=\frac{3±9}{2} when ± is plus and when ± is minus.

$$x=6 x=-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-6\right)\left(x+3\right)>0$$

For the product to be positive, x-6 and x+3 have to be both negative or both positive. Consider the case when x-6 and x+3 are both negative.

$$x-6<0 x+3<0$$

The solution satisfying both inequalities is x<-3.

$$x<-3$$

Consider the case when x-6 and x+3 are both positive.

$$x+3>0 x-6>0$$

The solution satisfying both inequalities is x>6.

$$x>6$$

The final solution is the union of the obtained solutions.

$$x<-3\text{; }x>6$$