Solve 4x^2+6x-1>0

$$4 x ^ { 2 } + 6 x - 1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{13}-3}{4}\right)\cup \left(\frac{\sqrt{13}-3}{4},\infty\right)$$

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.

$$4x^{2}+6x-1=0$$

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 4 for a, 6 for b, and -1 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-6±2\sqrt{13}}{8}$$

Solve the equation x=\frac{-6±2\sqrt{13}}{8} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{13}-3}{4} x=\frac{-\sqrt{13}-3}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{\sqrt{13}-3}{4}\right)\left(x-\frac{-\sqrt{13}-3}{4}\right)>0$$

For the product to be positive, x-\frac{\sqrt{13}-3}{4} and x-\frac{-\sqrt{13}-3}{4} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{13}-3}{4} and x-\frac{-\sqrt{13}-3}{4} are both negative.

$$x-\frac{\sqrt{13}-3}{4}<0 x-\frac{-\sqrt{13}-3}{4}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{13}-3}{4}.

$$x<\frac{-\sqrt{13}-3}{4}$$

Consider the case when x-\frac{\sqrt{13}-3}{4} and x-\frac{-\sqrt{13}-3}{4} are both positive.

$$x-\frac{-\sqrt{13}-3}{4}>0 x-\frac{\sqrt{13}-3}{4}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{13}-3}{4}.

$$x>\frac{\sqrt{13}-3}{4}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{13}-3}{4}\text{; }x>\frac{\sqrt{13}-3}{4}$$