Solve -6x^2-x+2geq0

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

Solve for x

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$

Steps Using the Quadratic Formula

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

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

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.

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

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 6 for a, 1 for b, and -2 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-1±7}{12}$$

Solve the equation x=\frac{-1±7}{12} when ± is plus and when ± is minus.

$$x=\frac{1}{2} x=-\frac{2}{3}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{1}{2} and x+\frac{2}{3} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{1}{2}\geq 0 and x+\frac{2}{3}\leq 0.

$$x-\frac{1}{2}\geq 0 x+\frac{2}{3}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{1}{2}\leq 0 and x+\frac{2}{3}\geq 0.

$$x+\frac{2}{3}\geq 0 x-\frac{1}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{2}{3},\frac{1}{2}\right].

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{2}{3},\frac{1}{2}\end{bmatrix}$$