Solve -m^2+12m-10<0

$$- { m }^{ 2 } +12m-10 < 0$$

Solve for m

$$m\in \left(-\infty,6-\sqrt{26}\right)\cup \left(\sqrt{26}+6,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -m^{2}+12m-10 positive. Since -1 is negative, the inequality direction is changed.

$$m^{2}-12m+10>0$$

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.

$$m^{2}-12m+10=0$$

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, -12 for b, and 10 for c in the quadratic formula.

$$m=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$m=\frac{12±2\sqrt{26}}{2}$$

Solve the equation m=\frac{12±2\sqrt{26}}{2} when ± is plus and when ± is minus.

$$m=\sqrt{26}+6 m=6-\sqrt{26}$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-\left(\sqrt{26}+6\right)\right)\left(m-\left(6-\sqrt{26}\right)\right)>0$$

For the product to be positive, m-\left(\sqrt{26}+6\right) and m-\left(6-\sqrt{26}\right) have to be both negative or both positive. Consider the case when m-\left(\sqrt{26}+6\right) and m-\left(6-\sqrt{26}\right) are both negative.

$$m-\left(\sqrt{26}+6\right)<0 m-\left(6-\sqrt{26}\right)<0$$

The solution satisfying both inequalities is m<6-\sqrt{26}.

$$m<6-\sqrt{26}$$

Consider the case when m-\left(\sqrt{26}+6\right) and m-\left(6-\sqrt{26}\right) are both positive.

$$m-\left(6-\sqrt{26}\right)>0 m-\left(\sqrt{26}+6\right)>0$$

The solution satisfying both inequalities is m>\sqrt{26}+6.

$$m>\sqrt{26}+6$$

The final solution is the union of the obtained solutions.

$$m<6-\sqrt{26}\text{; }m>\sqrt{26}+6$$