Solve 20x^2-23x+6geq0

$$20 x ^ { 2 } - 23 x + 6 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{2}{5}]\cup [\frac{3}{4},\infty)$$

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.

$$20x^{2}-23x+6=0$$

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 20 for a, -23 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 20\times 6}}{2\times 20}$$

Do the calculations.

$$x=\frac{23±7}{40}$$

Solve the equation x=\frac{23±7}{40} when ± is plus and when ± is minus.

$$x=\frac{3}{4} x=\frac{2}{5}$$

Rewrite the inequality by using the obtained solutions.

$$20\left(x-\frac{3}{4}\right)\left(x-\frac{2}{5}\right)\geq 0$$

For the product to be ≥0, x-\frac{3}{4} and x-\frac{2}{5} have to be both ≤0 or both ≥0. Consider the case when x-\frac{3}{4} and x-\frac{2}{5} are both ≤0.

$$x-\frac{3}{4}\leq 0 x-\frac{2}{5}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{2}{5}.

$$x\leq \frac{2}{5}$$

Consider the case when x-\frac{3}{4} and x-\frac{2}{5} are both ≥0.

$$x-\frac{2}{5}\geq 0 x-\frac{3}{4}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{3}{4}.

$$x\geq \frac{3}{4}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{2}{5}\text{; }x\geq \frac{3}{4}$$