Solve -3x^2+12x+6geq0

$$- 3 x ^ { 2 } + 12 x + 6 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}2-\sqrt{6},\sqrt{6}+2\end{bmatrix}$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -3x^{2}+12x+6 positive. Since -1 is negative, the inequality direction is changed.

$$3x^{2}-12x-6\leq 0$$

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.

$$3x^{2}-12x-6=0$$

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, -12 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{12±6\sqrt{6}}{6}$$

Solve the equation x=\frac{12±6\sqrt{6}}{6} when ± is plus and when ± is minus.

$$x=\sqrt{6}+2 x=2-\sqrt{6}$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be ≤0, one of the values x-\left(\sqrt{6}+2\right) and x-\left(2-\sqrt{6}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\sqrt{6}+2\right)\geq 0 and x-\left(2-\sqrt{6}\right)\leq 0.

$$x-\left(\sqrt{6}+2\right)\geq 0 x-\left(2-\sqrt{6}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\sqrt{6}+2\right)\leq 0 and x-\left(2-\sqrt{6}\right)\geq 0.

$$x-\left(2-\sqrt{6}\right)\geq 0 x-\left(\sqrt{6}+2\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[2-\sqrt{6},\sqrt{6}+2\right].

$$x\in \begin{bmatrix}2-\sqrt{6},\sqrt{6}+2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}2-\sqrt{6},\sqrt{6}+2\end{bmatrix}$$