Solve 3(x-8)(2-4x)leq0

$$3(x-8)(2-4x) \leq 0$$

Solve for x

$$x\in (-\infty,\frac{1}{2}]\cup [8,\infty)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-8.

$$\left(3x-24\right)\left(2-4x\right)\leq 0$$

Use the distributive property to multiply 3x-24 by 2-4x and combine like terms.

$$102x-12x^{2}-48\leq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 102x-12x^{2}-48 positive. Since -1 is negative, the inequality direction is changed.

$$-102x+12x^{2}+48\geq 0$$

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.

$$-102x+12x^{2}+48=0$$

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 12 for a, -102 for b, and 48 for c in the quadratic formula.

$$x=\frac{-\left(-102\right)±\sqrt{\left(-102\right)^{2}-4\times 12\times 48}}{2\times 12}$$

Do the calculations.

$$x=\frac{102±90}{24}$$

Solve the equation x=\frac{102±90}{24} when ± is plus and when ± is minus.

$$x=8 x=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$12\left(x-8\right)\left(x-\frac{1}{2}\right)\geq 0$$

For the product to be ≥0, x-8 and x-\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when x-8 and x-\frac{1}{2} are both ≤0.

$$x-8\leq 0 x-\frac{1}{2}\leq 0$$

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

$$x\leq \frac{1}{2}$$

Consider the case when x-8 and x-\frac{1}{2} are both ≥0.

$$x-\frac{1}{2}\geq 0 x-8\geq 0$$

The solution satisfying both inequalities is x\geq 8.

$$x\geq 8$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{1}{2}\text{; }x\geq 8$$