Solve (2)3x^2-8x+4>0

$$( 2 ) 3 x ^ { 2 } - 8 x + 4 > 0$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Multiply 2 and 3 to get 6.

$$6x^{2}-8x+4>0$$

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.

$$6x^{2}-8x+4=0$$

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

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

Do the calculations.

$$x=\frac{8±\sqrt{-32}}{12}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 6x^{2}-8x+4 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$6\times 0^{2}-8\times 0+4=4$$

The value of the expression 6x^{2}-8x+4 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$