Solve 6x^2-12x+3geq0

$$6 x ^ { 2 } - 12 x + 3 \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{\sqrt{2}}{2}+1]\cup [\frac{\sqrt{2}}{2}+1,\infty)$$

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.

$$6x^{2}-12x+3=0$$

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, -12 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\times 3}}{2\times 6}$$

Do the calculations.

$$x=\frac{12±6\sqrt{2}}{12}$$

Solve the equation x=\frac{12±6\sqrt{2}}{12} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{2}}{2}+1 x=-\frac{\sqrt{2}}{2}+1$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\left(\frac{\sqrt{2}}{2}+1\right)\right)\left(x-\left(-\frac{\sqrt{2}}{2}+1\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\frac{\sqrt{2}}{2}+1\right) and x-\left(-\frac{\sqrt{2}}{2}+1\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\frac{\sqrt{2}}{2}+1\right) and x-\left(-\frac{\sqrt{2}}{2}+1\right) are both ≤0.

$$x-\left(\frac{\sqrt{2}}{2}+1\right)\leq 0 x-\left(-\frac{\sqrt{2}}{2}+1\right)\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{\sqrt{2}}{2}+1.

$$x\leq -\frac{\sqrt{2}}{2}+1$$

Consider the case when x-\left(\frac{\sqrt{2}}{2}+1\right) and x-\left(-\frac{\sqrt{2}}{2}+1\right) are both ≥0.

$$x-\left(-\frac{\sqrt{2}}{2}+1\right)\geq 0 x-\left(\frac{\sqrt{2}}{2}+1\right)\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{2}}{2}+1.

$$x\geq \frac{\sqrt{2}}{2}+1$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{\sqrt{2}}{2}+1\text{; }x\geq \frac{\sqrt{2}}{2}+1$$