Solve 8y^2+4y-1>=0

$$8 { y }^{ 2 } +4y-1 > =0$$

Solve for y

$$y\in \left(-\infty,\frac{-\sqrt{3}-1}{4}\right)\cup \left(\frac{\sqrt{3}-1}{4},\infty\right)$$

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.

$$8y^{2}+4y-1=0$$

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

$$y=\frac{-4±\sqrt{4^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Do the calculations.

$$y=\frac{-4±4\sqrt{3}}{16}$$

Solve the equation y=\frac{-4±4\sqrt{3}}{16} when ± is plus and when ± is minus.

$$y=\frac{\sqrt{3}-1}{4} y=\frac{-\sqrt{3}-1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(y-\frac{\sqrt{3}-1}{4}\right)\left(y-\frac{-\sqrt{3}-1}{4}\right)>0$$

For the product to be positive, y-\frac{\sqrt{3}-1}{4} and y-\frac{-\sqrt{3}-1}{4} have to be both negative or both positive. Consider the case when y-\frac{\sqrt{3}-1}{4} and y-\frac{-\sqrt{3}-1}{4} are both negative.

$$y-\frac{\sqrt{3}-1}{4}<0 y-\frac{-\sqrt{3}-1}{4}<0$$

The solution satisfying both inequalities is y<\frac{-\sqrt{3}-1}{4}.

$$y<\frac{-\sqrt{3}-1}{4}$$

Consider the case when y-\frac{\sqrt{3}-1}{4} and y-\frac{-\sqrt{3}-1}{4} are both positive.

$$y-\frac{-\sqrt{3}-1}{4}>0 y-\frac{\sqrt{3}-1}{4}>0$$

The solution satisfying both inequalities is y>\frac{\sqrt{3}-1}{4}.

$$y>\frac{\sqrt{3}-1}{4}$$

The final solution is the union of the obtained solutions.

$$y<\frac{-\sqrt{3}-1}{4}\text{; }y>\frac{\sqrt{3}-1}{4}$$