Solve m^2+m-2>0

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

Solve for m

$$m\in \left(-\infty,-2\right)\cup \left(1,\infty\right)$$

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.

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

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, 1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{-1±3}{2}$$

Solve the equation m=\frac{-1±3}{2} when ± is plus and when ± is minus.

$$m=1 m=-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-1\right)\left(m+2\right)>0$$

For the product to be positive, m-1 and m+2 have to be both negative or both positive. Consider the case when m-1 and m+2 are both negative.

$$m-1<0 m+2<0$$

The solution satisfying both inequalities is m<-2.

$$m<-2$$

Consider the case when m-1 and m+2 are both positive.

$$m+2>0 m-1>0$$

The solution satisfying both inequalities is m>1.

$$m>1$$

The final solution is the union of the obtained solutions.

$$m<-2\text{; }m>1$$