Solve -3m^2-2m+1geq0

$$- 3 m ^ { 2 } - 2 m + 1 \geq 0$$

Solve for m

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

Steps Using the Quadratic Formula

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

$$3m^{2}+2m-1\leq 0$$

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.

$$3m^{2}+2m-1=0$$

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

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

Do the calculations.

$$m=\frac{-2±4}{6}$$

Solve the equation m=\frac{-2±4}{6} when ± is plus and when ± is minus.

$$m=\frac{1}{3} m=-1$$

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any m.

$$m\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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