Solve -x^2-6x+9<0

$$- x ^ { 2 } - 6 x + 9 < 0$$

Solve for x

$$x\in \left(-\infty,-3\sqrt{2}-3\right)\cup \left(3\sqrt{2}-3,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}-6x+9 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}+6x-9>0$$

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.

$$x^{2}+6x-9=0$$

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, 6 for b, and -9 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{-6±6\sqrt{2}}{2}$$

Solve the equation x=\frac{-6±6\sqrt{2}}{2} when ± is plus and when ± is minus.

$$x=3\sqrt{2}-3 x=-3\sqrt{2}-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(3\sqrt{2}-3\right)\right)\left(x-\left(-3\sqrt{2}-3\right)\right)>0$$

For the product to be positive, x-\left(3\sqrt{2}-3\right) and x-\left(-3\sqrt{2}-3\right) have to be both negative or both positive. Consider the case when x-\left(3\sqrt{2}-3\right) and x-\left(-3\sqrt{2}-3\right) are both negative.

$$x-\left(3\sqrt{2}-3\right)<0 x-\left(-3\sqrt{2}-3\right)<0$$

The solution satisfying both inequalities is x<-3\sqrt{2}-3.

$$x<-3\sqrt{2}-3$$

Consider the case when x-\left(3\sqrt{2}-3\right) and x-\left(-3\sqrt{2}-3\right) are both positive.

$$x-\left(-3\sqrt{2}-3\right)>0 x-\left(3\sqrt{2}-3\right)>0$$

The solution satisfying both inequalities is x>3\sqrt{2}-3.

$$x>3\sqrt{2}-3$$

The final solution is the union of the obtained solutions.

$$x<-3\sqrt{2}-3\text{; }x>3\sqrt{2}-3$$