Solve {l}{m^2-2m}{-3>0}

$$\left. \begin{array} { l } { m ^ { 2 } - 2 m } \\ { - 3 > 0 } \end{array} \right.$$

Solve for m

$$m\in \left(-\infty,-1\right)\cup \left(3,\infty\right)$$

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.

$$m^{2}-2m-3=0$$

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, -2 for b, and -3 for c in the quadratic formula.

$$m=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$m=\frac{2±4}{2}$$

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

$$m=3 m=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-3\right)\left(m+1\right)>0$$

For the product to be positive, m-3 and m+1 have to be both negative or both positive. Consider the case when m-3 and m+1 are both negative.

$$m-3<0 m+1<0$$

The solution satisfying both inequalities is m<-1.

$$m<-1$$

Consider the case when m-3 and m+1 are both positive.

$$m+1>0 m-3>0$$

The solution satisfying both inequalities is m>3.

$$m>3$$

The final solution is the union of the obtained solutions.

$$m<-1\text{; }m>3$$