Solve -8a^2+6a-1>0

$$- 8 a ^ { 2 } + 6 a - 1 > 0$$

Solve for a

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -8a^{2}+6a-1 positive. Since -1 is negative, the inequality direction is changed.

$$8a^{2}-6a+1<0$$

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.

$$8a^{2}-6a+1=0$$

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

$$a=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\times 1}}{2\times 8}$$

Do the calculations.

$$a=\frac{6±2}{16}$$

Solve the equation a=\frac{6±2}{16} when ± is plus and when ± is minus.

$$a=\frac{1}{2} a=\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(a-\frac{1}{2}\right)\left(a-\frac{1}{4}\right)<0$$

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

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

This is false for any a.

$$a\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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