Solve 2x^2+3x-4<0

$$2 { x }^{ 2 } +3x-4 < 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{41}-3}{4},\frac{\sqrt{41}-3}{4}\right)$$

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.

$$2x^{2}+3x-4=0$$

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

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

Do the calculations.

$$x=\frac{-3±\sqrt{41}}{4}$$

Solve the equation x=\frac{-3±\sqrt{41}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{41}-3}{4}\right)\left(x-\frac{-\sqrt{41}-3}{4}\right)<0$$

For the product to be negative, x-\frac{\sqrt{41}-3}{4} and x-\frac{-\sqrt{41}-3}{4} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{41}-3}{4} is positive and x-\frac{-\sqrt{41}-3}{4} is negative.

$$x-\frac{\sqrt{41}-3}{4}>0 x-\frac{-\sqrt{41}-3}{4}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{41}-3}{4} is positive and x-\frac{\sqrt{41}-3}{4} is negative.

$$x-\frac{-\sqrt{41}-3}{4}>0 x-\frac{\sqrt{41}-3}{4}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{41}-3}{4},\frac{\sqrt{41}-3}{4}\right).

$$x\in \left(\frac{-\sqrt{41}-3}{4},\frac{\sqrt{41}-3}{4}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{41}-3}{4},\frac{\sqrt{41}-3}{4}\right)$$