Solve 2m^2-m-1>0

$$2 m ^ { 2 } - m - 1 > 0$$

Solve for m

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

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.

$$2m^{2}-m-1=0$$

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

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

Do the calculations.

$$m=\frac{1±3}{4}$$

Solve the equation m=\frac{1±3}{4} when ± is plus and when ± is minus.

$$m=1 m=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

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

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

$$m-1<0 m+\frac{1}{2}<0$$

The solution satisfying both inequalities is m<-\frac{1}{2}.

$$m<-\frac{1}{2}$$

Consider the case when m-1 and m+\frac{1}{2} are both positive.

$$m+\frac{1}{2}>0 m-1>0$$

The solution satisfying both inequalities is m>1.

$$m>1$$

The final solution is the union of the obtained solutions.

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