Solve a^2-2a-3>0

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

Solve for a

$$a\in \left(-\infty,-1\right)\cup \left(3,\infty\right)$$

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.

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

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, -2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{2±4}{2}$$

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

$$a=3 a=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-3\right)\left(a+1\right)>0$$

For the product to be positive, a-3 and a+1 have to be both negative or both positive. Consider the case when a-3 and a+1 are both negative.

$$a-3<0 a+1<0$$

The solution satisfying both inequalities is a<-1.

$$a<-1$$

Consider the case when a-3 and a+1 are both positive.

$$a+1>0 a-3>0$$

The solution satisfying both inequalities is a>3.

$$a>3$$

The final solution is the union of the obtained solutions.

$$a<-1\text{; }a>3$$