Solve x^2-16x+60>0

$$x ^ { 2 } - 16 x + 60 > 0$$

Solve for x

$$x\in \left(-\infty,6\right)\cup \left(10,\infty\right)$$

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.

$$x^{2}-16x+60=0$$

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, -16 for b, and 60 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\times 60}}{2}$$

Do the calculations.

$$x=\frac{16±4}{2}$$

Solve the equation x=\frac{16±4}{2} when ± is plus and when ± is minus.

$$x=10 x=6$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-10\right)\left(x-6\right)>0$$

For the product to be positive, x-10 and x-6 have to be both negative or both positive. Consider the case when x-10 and x-6 are both negative.

$$x-10<0 x-6<0$$

The solution satisfying both inequalities is x<6.

$$x<6$$

Consider the case when x-10 and x-6 are both positive.

$$x-6>0 x-10>0$$

The solution satisfying both inequalities is x>10.

$$x>10$$

The final solution is the union of the obtained solutions.

$$x<6\text{; }x>10$$