Solve 2x^2-7x-4<0

$$2 { x }^{ 2 } -7x-4 < 0$$

Solve for x

$$x\in \left(-\frac{1}{2},4\right)$$

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.

$$2x^{2}-7x-4=0$$

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, -7 for b, and -4 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{7±9}{4}$$

Solve the equation x=\frac{7±9}{4} when ± is plus and when ± is minus.

$$x=4 x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-4\right)\left(x+\frac{1}{2}\right)<0$$

For the product to be negative, x-4 and x+\frac{1}{2} have to be of the opposite signs. Consider the case when x-4 is positive and x+\frac{1}{2} is negative.

$$x-4>0 x+\frac{1}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{2} is positive and x-4 is negative.

$$x+\frac{1}{2}>0 x-4<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{2},4\right).

$$x\in \left(-\frac{1}{2},4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{1}{2},4\right)$$