Solve -2x^2+2x+12leq0

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

Solve for x

$$x\in (-\infty,-2]\cup [3,\infty)$$

Steps Using the Quadratic Formula

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

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

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.

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

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 2 for a, -2 for b, and -12 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±10}{4}$$

Solve the equation x=\frac{2±10}{4} when ± is plus and when ± is minus.

$$x=3 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-3\right)\left(x+2\right)\geq 0$$

For the product to be ≥0, x-3 and x+2 have to be both ≤0 or both ≥0. Consider the case when x-3 and x+2 are both ≤0.

$$x-3\leq 0 x+2\leq 0$$

The solution satisfying both inequalities is x\leq -2.

$$x\leq -2$$

Consider the case when x-3 and x+2 are both ≥0.

$$x+2\geq 0 x-3\geq 0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq -2\text{; }x\geq 3$$