Solve x^2-17x+72>0

$${ x }^{ 2 } -17x+72 > 0$$

Solve for x

$$x\in \left(-\infty,8\right)\cup \left(9,\infty\right)$$

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.

$$x^{2}-17x+72=0$$

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, -17 for b, and 72 for c in the quadratic formula.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 1\times 72}}{2}$$

Do the calculations.

$$x=\frac{17±1}{2}$$

Solve the equation x=\frac{17±1}{2} when ± is plus and when ± is minus.

$$x=9 x=8$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-9\right)\left(x-8\right)>0$$

For the product to be positive, x-9 and x-8 have to be both negative or both positive. Consider the case when x-9 and x-8 are both negative.

$$x-9<0 x-8<0$$

The solution satisfying both inequalities is x<8.

$$x<8$$

Consider the case when x-9 and x-8 are both positive.

$$x-8>0 x-9>0$$

The solution satisfying both inequalities is x>9.

$$x>9$$

The final solution is the union of the obtained solutions.

$$x<8\text{; }x>9$$