Solve 3x^2-8x+4<0

$$3 x ^ { 2 } - 8 x + 4 < 0$$

Solve for x

$$x\in \left(\frac{2}{3},2\right)$$

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.

$$3x^{2}-8x+4=0$$

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

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Do the calculations.

$$x=\frac{8±4}{6}$$

Solve the equation x=\frac{8±4}{6} when ± is plus and when ± is minus.

$$x=2 x=\frac{2}{3}$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be negative, x-2 and x-\frac{2}{3} have to be of the opposite signs. Consider the case when x-2 is positive and x-\frac{2}{3} is negative.

$$x-2>0 x-\frac{2}{3}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{2}{3} is positive and x-2 is negative.

$$x-\frac{2}{3}>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(\frac{2}{3},2\right).

$$x\in \left(\frac{2}{3},2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{2}{3},2\right)$$