Solve x^2+13x+40geq0

$$x ^ { 2 } + 13 x + 40 \geq 0$$

Solve for x

$$x\in (-\infty,-8]\cup [-5,\infty)$$

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.

$$x^{2}+13x+40=0$$

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, 13 for b, and 40 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 1\times 40}}{2}$$

Do the calculations.

$$x=\frac{-13±3}{2}$$

Solve the equation x=\frac{-13±3}{2} when ± is plus and when ± is minus.

$$x=-5 x=-8$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+5\right)\left(x+8\right)\geq 0$$

For the product to be ≥0, x+5 and x+8 have to be both ≤0 or both ≥0. Consider the case when x+5 and x+8 are both ≤0.

$$x+5\leq 0 x+8\leq 0$$

The solution satisfying both inequalities is x\leq -8.

$$x\leq -8$$

Consider the case when x+5 and x+8 are both ≥0.

$$x+8\geq 0 x+5\geq 0$$

The solution satisfying both inequalities is x\geq -5.

$$x\geq -5$$

The final solution is the union of the obtained solutions.

$$x\leq -8\text{; }x\geq -5$$