Solve x^2+2x-3<0

$${ x }^{ 2 } +2x-3 < 0$$

Solve for x

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

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.

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

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.

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

Do the calculations.

$$x=\frac{-2±4}{2}$$

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

$$x=1 x=-3$$

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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