Solve x^2+2x-3>-x^2

$${ x }^{ 2 } +2x-3 > - { x }^{ 2 }$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{7}-1}{2}\right)\cup \left(\frac{\sqrt{7}-1}{2},\infty\right)$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

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.

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

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

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

Do the calculations.

$$x=\frac{-2±2\sqrt{7}}{4}$$

Solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{7}-1}{2} x=\frac{-\sqrt{7}-1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{7}-1}{2}\right)\left(x-\frac{-\sqrt{7}-1}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} are both negative.

$$x-\frac{\sqrt{7}-1}{2}<0 x-\frac{-\sqrt{7}-1}{2}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{7}-1}{2}.

$$x<\frac{-\sqrt{7}-1}{2}$$

Consider the case when x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} are both positive.

$$x-\frac{-\sqrt{7}-1}{2}>0 x-\frac{\sqrt{7}-1}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{7}-1}{2}.

$$x>\frac{\sqrt{7}-1}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{7}-1}{2}\text{; }x>\frac{\sqrt{7}-1}{2}$$