Solve (m-4)^2-4(6-m)>0

$$( m - 4 ) ^ { 2 } - 4 ( 6 - m ) > 0$$

Solve for m

$$m\in \left(-\infty,2-2\sqrt{3}\right)\cup \left(2\sqrt{3}+2,\infty\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-4\right)^{2}.

$$m^{2}-8m+16-4\left(6-m\right)>0$$

Use the distributive property to multiply -4 by 6-m.

$$m^{2}-8m+16-24+4m>0$$

Subtract 24 from 16 to get -8.

$$m^{2}-8m-8+4m>0$$

Combine -8m and 4m to get -4m.

$$m^{2}-4m-8>0$$

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.

$$m^{2}-4m-8=0$$

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, -4 for b, and -8 for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{4±4\sqrt{3}}{2}$$

Solve the equation m=\frac{4±4\sqrt{3}}{2} when ± is plus and when ± is minus.

$$m=2\sqrt{3}+2 m=2-2\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-\left(2\sqrt{3}+2\right)\right)\left(m-\left(2-2\sqrt{3}\right)\right)>0$$

For the product to be positive, m-\left(2\sqrt{3}+2\right) and m-\left(2-2\sqrt{3}\right) have to be both negative or both positive. Consider the case when m-\left(2\sqrt{3}+2\right) and m-\left(2-2\sqrt{3}\right) are both negative.

$$m-\left(2\sqrt{3}+2\right)<0 m-\left(2-2\sqrt{3}\right)<0$$

The solution satisfying both inequalities is m<2-2\sqrt{3}.

$$m<2-2\sqrt{3}$$

Consider the case when m-\left(2\sqrt{3}+2\right) and m-\left(2-2\sqrt{3}\right) are both positive.

$$m-\left(2-2\sqrt{3}\right)>0 m-\left(2\sqrt{3}+2\right)>0$$

The solution satisfying both inequalities is m>2\sqrt{3}+2.

$$m>2\sqrt{3}+2$$

The final solution is the union of the obtained solutions.

$$m<2-2\sqrt{3}\text{; }m>2\sqrt{3}+2$$