Solve 2x^2-3x-14>0

$$2 x ^ { 2 } - 3 x - 14 > 0$$

Solve for x

$$x\in \left(-\infty,-2\right)\cup \left(\frac{7}{2},\infty\right)$$

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.

$$2x^{2}-3x-14=0$$

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, -3 for b, and -14 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-14\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±11}{4}$$

Solve the equation x=\frac{3±11}{4} when ± is plus and when ± is minus.

$$x=\frac{7}{2} x=-2$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{7}{2}\right)\left(x+2\right)>0$$

For the product to be positive, x-\frac{7}{2} and x+2 have to be both negative or both positive. Consider the case when x-\frac{7}{2} and x+2 are both negative.

$$x-\frac{7}{2}<0 x+2<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when x-\frac{7}{2} and x+2 are both positive.

$$x+2>0 x-\frac{7}{2}>0$$

The solution satisfying both inequalities is x>\frac{7}{2}.

$$x>\frac{7}{2}$$

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x>\frac{7}{2}$$