Solve x^2+12x+20geq0

$${ x }^{ 2 } +12x+20 \geq 0$$

Solve for x

$$x\in (-\infty,-10]\cup [-2,\infty)$$

Steps Using the Quadratic Formula

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.

$$x^{2}+12x+20=0$$

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, 12 for b, and 20 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 1\times 20}}{2}$$

Do the calculations.

$$x=\frac{-12±8}{2}$$

Solve the equation x=\frac{-12±8}{2} when ± is plus and when ± is minus.

$$x=-2 x=-10$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+2\right)\left(x+10\right)\geq 0$$

For the product to be ≥0, x+2 and x+10 have to be both ≤0 or both ≥0. Consider the case when x+2 and x+10 are both ≤0.

$$x+2\leq 0 x+10\leq 0$$

The solution satisfying both inequalities is x\leq -10.

$$x\leq -10$$

Consider the case when x+2 and x+10 are both ≥0.

$$x+10\geq 0 x+2\geq 0$$

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

$$x\geq -2$$

The final solution is the union of the obtained solutions.

$$x\leq -10\text{; }x\geq -2$$