Solve 0<8x^2-20x+8

$$0 < 8 { x }^{ 2 } -20x+8$$

Solve for x

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

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$8x^{2}-20x+8>0$$

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.

$$8x^{2}-20x+8=0$$

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, -20 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 8\times 8}}{2\times 8}$$

Do the calculations.

$$x=\frac{20±12}{16}$$

Solve the equation x=\frac{20±12}{16} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$8\left(x-2\right)\left(x-\frac{1}{2}\right)>0$$

For the product to be positive, x-2 and x-\frac{1}{2} have to be both negative or both positive. Consider the case when x-2 and x-\frac{1}{2} are both negative.

$$x-2<0 x-\frac{1}{2}<0$$

The solution satisfying both inequalities is x<\frac{1}{2}.

$$x<\frac{1}{2}$$

Consider the case when x-2 and x-\frac{1}{2} are both positive.

$$x-\frac{1}{2}>0 x-2>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{2}\text{; }x>2$$