Solve x^2-8x-9>0

$$x ^ { 2 } - 8 x - 9 > 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(9,\infty\right)$$

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.

$$x^{2}-8x-9=0$$

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, -8 for b, and -9 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{8±10}{2}$$

Solve the equation x=\frac{8±10}{2} when ± is plus and when ± is minus.

$$x=9 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-9\right)\left(x+1\right)>0$$

For the product to be positive, x-9 and x+1 have to be both negative or both positive. Consider the case when x-9 and x+1 are both negative.

$$x-9<0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x-9 and x+1 are both positive.

$$x+1>0 x-9>0$$

The solution satisfying both inequalities is x>9.

$$x>9$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>9$$