Solve 2x^2-8x+3geq0

$$2 x ^ { 2 } - 8 x + 3 \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{\sqrt{10}}{2}+2]\cup [\frac{\sqrt{10}}{2}+2,\infty)$$

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.

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

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

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

Do the calculations.

$$x=\frac{8±2\sqrt{10}}{4}$$

Solve the equation x=\frac{8±2\sqrt{10}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{10}}{2}+2 x=-\frac{\sqrt{10}}{2}+2$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\left(\frac{\sqrt{10}}{2}+2\right)\right)\left(x-\left(-\frac{\sqrt{10}}{2}+2\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\frac{\sqrt{10}}{2}+2\right) and x-\left(-\frac{\sqrt{10}}{2}+2\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\frac{\sqrt{10}}{2}+2\right) and x-\left(-\frac{\sqrt{10}}{2}+2\right) are both ≤0.

$$x-\left(\frac{\sqrt{10}}{2}+2\right)\leq 0 x-\left(-\frac{\sqrt{10}}{2}+2\right)\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{\sqrt{10}}{2}+2.

$$x\leq -\frac{\sqrt{10}}{2}+2$$

Consider the case when x-\left(\frac{\sqrt{10}}{2}+2\right) and x-\left(-\frac{\sqrt{10}}{2}+2\right) are both ≥0.

$$x-\left(-\frac{\sqrt{10}}{2}+2\right)\geq 0 x-\left(\frac{\sqrt{10}}{2}+2\right)\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{10}}{2}+2.

$$x\geq \frac{\sqrt{10}}{2}+2$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{\sqrt{10}}{2}+2\text{; }x\geq \frac{\sqrt{10}}{2}+2$$