Solve 4m^2-12(m+6)>0

$$4 m ^ { 2 } - 12 ( m + 6 ) > 0$$

Solve for m

$$m\in \left(-\infty,-3\right)\cup \left(6,\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -12 by m+6.

$$4m^{2}-12m-72>0$$

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.

$$4m^{2}-12m-72=0$$

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 4 for a, -12 for b, and -72 for c in the quadratic formula.

$$m=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-72\right)}}{2\times 4}$$

Do the calculations.

$$m=\frac{12±36}{8}$$

Solve the equation m=\frac{12±36}{8} when ± is plus and when ± is minus.

$$m=6 m=-3$$

Rewrite the inequality by using the obtained solutions.

$$4\left(m-6\right)\left(m+3\right)>0$$

For the product to be positive, m-6 and m+3 have to be both negative or both positive. Consider the case when m-6 and m+3 are both negative.

$$m-6<0 m+3<0$$

The solution satisfying both inequalities is m<-3.

$$m<-3$$

Consider the case when m-6 and m+3 are both positive.

$$m+3>0 m-6>0$$

The solution satisfying both inequalities is m>6.

$$m>6$$

The final solution is the union of the obtained solutions.

$$m<-3\text{; }m>6$$