Solve {l}{-6x^2-x}{+2leq0}

$$\left. \begin{array} { l } { - 6 x ^ { 2 } - x } \\ { + 2 \leq 0 } \end{array} \right.$$

Solve for x

$$x\in (-\infty,-\frac{2}{3}]\cup [\frac{1}{2},\infty)$$

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.

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

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.

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

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.

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

For the product to be ≥0, x-\frac{1}{2} and x+\frac{2}{3} have to be both ≤0 or both ≥0. Consider the case when x-\frac{1}{2} and x+\frac{2}{3} are both ≤0.

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

The solution satisfying both inequalities is x\leq -\frac{2}{3}.

$$x\leq -\frac{2}{3}$$

Consider the case when x-\frac{1}{2} and x+\frac{2}{3} are both ≥0.

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

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

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

The final solution is the union of the obtained solutions.

$$x\leq -\frac{2}{3}\text{; }x\geq \frac{1}{2}$$