Solve t^2-2t-3>0

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

Solve for t

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

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.

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

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.

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

Do the calculations.

$$t=\frac{2±4}{2}$$

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

$$t=3 t=-1$$

Rewrite the inequality by using the obtained solutions.

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

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

$$t-3<0 t+1<0$$

The solution satisfying both inequalities is t<-1.

$$t<-1$$

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

$$t+1>0 t-3>0$$

The solution satisfying both inequalities is t>3.

$$t>3$$

The final solution is the union of the obtained solutions.

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