Solve m^2+4m-4>0

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

Solve for m

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

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.

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

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 -4 for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{-4±4\sqrt{2}}{2}$$

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

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

Rewrite the inequality by using the obtained solutions.

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

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

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

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

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

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

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

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

$$m>2\sqrt{2}-2$$

The final solution is the union of the obtained solutions.

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