Solve -x^2-4x+12geq0

$$- x ^ { 2 } - 4 x + 12 \geq 0$$

Solve for x

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

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -x^{2}-4x+12 positive. Since -1 is negative, the inequality direction is changed.

$$x^{2}+4x-12\leq 0$$

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.

$$x^{2}+4x-12=0$$

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 1 for a, 4 for b, and -12 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{-4±8}{2}$$

Solve the equation x=\frac{-4±8}{2} when ± is plus and when ± is minus.

$$x=2 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-2\right)\left(x+6\right)\leq 0$$

For the product to be ≤0, one of the values x-2 and x+6 has to be ≥0 and the other has to be ≤0. Consider the case when x-2\geq 0 and x+6\leq 0.

$$x-2\geq 0 x+6\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-2\leq 0 and x+6\geq 0.

$$x+6\geq 0 x-2\leq 0$$

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

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

The final solution is the union of the obtained solutions.

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