Solve s^2-2s-3<0

$$s ^ { 2 } - 2 s - 3 < 0$$

Solve for s

$$s\in \left(-1,3\right)$$

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.

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

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.

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

Do the calculations.

$$s=\frac{2±4}{2}$$

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

$$s=3 s=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(s-3\right)\left(s+1\right)<0$$

For the product to be negative, s-3 and s+1 have to be of the opposite signs. Consider the case when s-3 is positive and s+1 is negative.

$$s-3>0 s+1<0$$

This is false for any s.

$$s\in \emptyset$$

Consider the case when s+1 is positive and s-3 is negative.

$$s+1>0 s-3<0$$

The solution satisfying both inequalities is s\in \left(-1,3\right).

$$s\in \left(-1,3\right)$$

The final solution is the union of the obtained solutions.

$$s\in \left(-1,3\right)$$