Solve m^2-8m+15>0

$${ m }^{ 2 } -8m+15 > 0$$

Solve for m

$$m\in \left(-\infty,3\right)\cup \left(5,\infty\right)$$

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.

$$m^{2}-8m+15=0$$

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 15 for c in the quadratic formula.

$$m=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 15}}{2}$$

Do the calculations.

$$m=\frac{8±2}{2}$$

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

$$m=5 m=3$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-5\right)\left(m-3\right)>0$$

For the product to be positive, m-5 and m-3 have to be both negative or both positive. Consider the case when m-5 and m-3 are both negative.

$$m-5<0 m-3<0$$

The solution satisfying both inequalities is m<3.

$$m<3$$

Consider the case when m-5 and m-3 are both positive.

$$m-3>0 m-5>0$$

The solution satisfying both inequalities is m>5.

$$m>5$$

The final solution is the union of the obtained solutions.

$$m<3\text{; }m>5$$