Solve -6x^2-18x+20geq0

$$- 6 x ^ { 2 } - 18 x + 20 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{\sqrt{201}}{6}-\frac{3}{2},\frac{\sqrt{201}}{6}-\frac{3}{2}\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -6x^{2}-18x+20 positive. Since -1 is negative, the inequality direction is changed.

$$6x^{2}+18x-20\leq 0$$

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.

$$6x^{2}+18x-20=0$$

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, 18 for b, and -20 for c in the quadratic formula.

$$x=\frac{-18±\sqrt{18^{2}-4\times 6\left(-20\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-18±2\sqrt{201}}{12}$$

Solve the equation x=\frac{-18±2\sqrt{201}}{12} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{201}}{6}-\frac{3}{2} x=-\frac{\sqrt{201}}{6}-\frac{3}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\right)\left(x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right) and x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\geq 0 and x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\leq 0.

$$x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\geq 0 x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\leq 0 and x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\geq 0.

$$x-\left(-\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\geq 0 x-\left(\frac{\sqrt{201}}{6}-\frac{3}{2}\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{\sqrt{201}}{6}-\frac{3}{2},\frac{\sqrt{201}}{6}-\frac{3}{2}\right].

$$x\in \begin{bmatrix}-\frac{\sqrt{201}}{6}-\frac{3}{2},\frac{\sqrt{201}}{6}-\frac{3}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{\sqrt{201}}{6}-\frac{3}{2},\frac{\sqrt{201}}{6}-\frac{3}{2}\end{bmatrix}$$