Answer
$$\color{blue}{(-\infty,-3]\cup[-1,\infty)}$$
Work Step by Step
$\bf\text{First, rewrite with all terms on the left side.}$
$-x^2-4x-3\leq{0}$
$\bf\text{Applying the quadratic formula gives us:}$
$x=-3$ or $x=-1$
$\bf\text{Which gives us three intervals:}$
$\bf(-\infty,-3]$ $\bf[-3,-1]$ $\bf[-1,\infty)$
$\bf\text{Using test values we can test each interval in the formula}$
For the interval $\bf(-\infty,-3]$, we can use $-4$ as our test value
$-(-4)^2-4(-4)-3\leq{0}$
$-16+16-3\leq{0}$
$-3\leq{0}$ which is true so this interval is $\bf\text{valid}$
For the interval $\bf[-3,-1]$ we can use $-2$ as our test value
$-(-2)^2-4(-2)-3\leq{0}$
$-4+8-3\leq{0}$
$1\leq{0}$ which is untrue so this interval is $\bf\text{not valid}$
For the interval $\bf[-1,\infty)$, we can use $1$ as our test value
$-(1)^2-4(1)-3\leq{0}$
$-1-4-3\leq{0}$
$-11\leq{0}$ which is true so this interval is $\bf\text{valid}$
Therefore the solution is $\color{blue}{(-\infty,-3]\cup[-1,\infty)}$
