Solve x^2-65x+900>0

$$x ^ { 2 } - 65 x + 900 > 0$$

Solve for x

$$x\in \left(-\infty,20\right)\cup \left(45,\infty\right)$$

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.

$$x^{2}-65x+900=0$$

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, -65 for b, and 900 for c in the quadratic formula.

$$x=\frac{-\left(-65\right)±\sqrt{\left(-65\right)^{2}-4\times 1\times 900}}{2}$$

Do the calculations.

$$x=\frac{65±25}{2}$$

Solve the equation x=\frac{65±25}{2} when ± is plus and when ± is minus.

$$x=45 x=20$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-45\right)\left(x-20\right)>0$$

For the product to be positive, x-45 and x-20 have to be both negative or both positive. Consider the case when x-45 and x-20 are both negative.

$$x-45<0 x-20<0$$

The solution satisfying both inequalities is x<20.

$$x<20$$

Consider the case when x-45 and x-20 are both positive.

$$x-20>0 x-45>0$$

The solution satisfying both inequalities is x>45.

$$x>45$$

The final solution is the union of the obtained solutions.

$$x<20\text{; }x>45$$