Solve a^2-3a-18>0

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

Solve for a

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

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.

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

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.

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

Do the calculations.

$$a=\frac{3±9}{2}$$

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

$$a=6 a=-3$$

Rewrite the inequality by using the obtained solutions.

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

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

$$a-6<0 a+3<0$$

The solution satisfying both inequalities is a<-3.

$$a<-3$$

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

$$a+3>0 a-6>0$$

The solution satisfying both inequalities is a>6.

$$a>6$$

The final solution is the union of the obtained solutions.

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