Solve -3x^2+x+2geq0

$$-3 { x }^{ 2 } +x+2 \geq 0$$

Solve for x

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -3x^{2}+x+2 positive. Since -1 is negative, the inequality direction is changed.

$$3x^{2}-x-2\leq 0$$

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.

$$3x^{2}-x-2=0$$

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

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 3\left(-2\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{1±5}{6}$$

Solve the equation x=\frac{1±5}{6} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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