Solve m^2+m-6<0

$$m ^ { 2 } + m - 6 < 0$$

Solve for m

$$m\in \left(-3,2\right)$$

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.

$$m^{2}+m-6=0$$

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 -6 for c in the quadratic formula.

$$m=\frac{-1±\sqrt{1^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$m=\frac{-1±5}{2}$$

Solve the equation m=\frac{-1±5}{2} when ± is plus and when ± is minus.

$$m=2 m=-3$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be negative, m-2 and m+3 have to be of the opposite signs. Consider the case when m-2 is positive and m+3 is negative.

$$m-2>0 m+3<0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m+3 is positive and m-2 is negative.

$$m+3>0 m-2<0$$

The solution satisfying both inequalities is m\in \left(-3,2\right).

$$m\in \left(-3,2\right)$$

The final solution is the union of the obtained solutions.

$$m\in \left(-3,2\right)$$