Solve -3k^2+6k+1geq0

$$- 3 k ^ { 2 } + 6 k + 1 \geq 0$$

Solve for k

$$k\in \begin{bmatrix}-\frac{2\sqrt{3}}{3}+1,\frac{2\sqrt{3}}{3}+1\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -3k^{2}+6k+1 positive. Since -1 is negative, the inequality direction is changed.

$$3k^{2}-6k-1\leq 0$$

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.

$$3k^{2}-6k-1=0$$

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

$$k=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$k=\frac{6±4\sqrt{3}}{6}$$

Solve the equation k=\frac{6±4\sqrt{3}}{6} when ± is plus and when ± is minus.

$$k=\frac{2\sqrt{3}}{3}+1 k=-\frac{2\sqrt{3}}{3}+1$$

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any k.

$$k\in \emptyset$$

Consider the case when k-\left(\frac{2\sqrt{3}}{3}+1\right)\leq 0 and k-\left(-\frac{2\sqrt{3}}{3}+1\right)\geq 0.

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

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

$$k\in \begin{bmatrix}-\frac{2\sqrt{3}}{3}+1,\frac{2\sqrt{3}}{3}+1\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$k\in \begin{bmatrix}-\frac{2\sqrt{3}}{3}+1,\frac{2\sqrt{3}}{3}+1\end{bmatrix}$$