Solve 1-2x-x^2>0

$$1 - 2 x - x ^ { 2 } > 0$$

Solve for x

$$x\in \left(-\sqrt{2}-1,\sqrt{2}-1\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 1-2x-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-1+2x+x^{2}<0$$

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.

$$-1+2x+x^{2}=0$$

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 -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{2}}{2}$$

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

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

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{2}-1\right)\right)\left(x-\left(-\sqrt{2}-1\right)\right)<0$$

For the product to be negative, x-\left(\sqrt{2}-1\right) and x-\left(-\sqrt{2}-1\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{2}-1\right) is positive and x-\left(-\sqrt{2}-1\right) is negative.

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

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(-\sqrt{2}-1\right) is positive and x-\left(\sqrt{2}-1\right) is negative.

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

The solution satisfying both inequalities is x\in \left(-\left(\sqrt{2}+1\right),\sqrt{2}-1\right).

$$x\in \left(-\left(\sqrt{2}+1\right),\sqrt{2}-1\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\sqrt{2}-1,\sqrt{2}-1\right)$$