Solve 7(x+0,6)(x-6)leq0

$$7 ( x + 0,6 ) ( x - 6 ) \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-0,6;6\end{bmatrix}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7 by x+0,6.

$$\left(7x+4,2\right)\left(x-6\right)\leq 0$$

Use the distributive property to multiply 7x+4,2 by x-6 and combine like terms.

$$7x^{2}-37,8x-25,2\leq 0$$

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.

$$7x^{2}-37,8x-25,2=0$$

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 7 for a, -37,8 for b, and -25,2 for c in the quadratic formula.

$$x=\frac{-\left(-37,8\right)±\sqrt{\left(-37,8\right)^{2}-4\times 7\left(-25,2\right)}}{2\times 7}$$

Do the calculations.

$$x=\frac{37,8±\frac{231}{5}}{14}$$

Solve the equation x=\frac{37,8±\frac{231}{5}}{14} when ± is plus and when ± is minus.

$$x=6 x=-\frac{3}{5}$$

Rewrite the inequality by using the obtained solutions.

$$7\left(x-6\right)\left(x+\frac{3}{5}\right)\leq 0$$

For the product to be ≤0, one of the values x-6 and x+\frac{3}{5} has to be ≥0 and the other has to be ≤0. Consider the case when x-6\geq 0 and x+\frac{3}{5}\leq 0.

$$x-6\geq 0 x+\frac{3}{5}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-6\leq 0 and x+\frac{3}{5}\geq 0.

$$x+\frac{3}{5}\geq 0 x-6\leq 0$$

The solution satisfying both inequalities is x\in \left[-0,6;6\right].

$$x\in \begin{bmatrix}-0,6;6\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{3}{5};6\end{bmatrix}$$