Question

determine the solution set of the quadratic inequality: $x^{2}+6x - 5geq - 14$. your answer should be in interval notation.
$(-infty,infty)$
$(-infty, - 3)cup(-3,infty)$
$(-3,3)$
$(-infty, - 3)cup(3,infty)$

Explanation:

Step1: Rearrange the inequality

Add 14 to both sides of $x^{2}+6x - 5\geq - 14$ to get $x^{2}+6x + 9\geq0$.

Step2: Factor the quadratic expression

The left - hand side $x^{2}+6x + 9$ can be factored as $(x + 3)^{2}\geq0$.

Step3: Analyze the non - negativity of the squared expression

Since the square of any real number is non - negative, i.e., for any real number $x$, $(x + 3)^{2}\geq0$ holds true. The solution set of the inequality is all real numbers.

Answer:

$(-\infty,\infty)$