jmc

After dividing both sides by 2 and moving the constant over, we get a quadratic expression and solve for the roots: $$\begin{array}{rl}{x}^2+4x+3& \le 0\Rightarrow \\ (x+1)(x+3)& \le 0.\end{array}$$The quadratic expression equals 0 at $x=-3$ and $x=-1$, meaning it changes sign at each root. Now we look at the sign of the quadratic when $x<-3$, when $-3<x<-1$, and when $x>-1$. When $x<-3$, $(x+3)$ and $(x+1)$ are both negative, so the product is positive. When $-3<x<-1$, $(x+3)$ becomes positive, while $(x+1)$ remains negative - the product is negative. When $x>-1$, both factors are positive, so the product is positive. So, $(x+1)(x+3)\le 0$ when $-3\le x\le -1$, which means our answer in interval notation is $\boxed{[-3,-1]}$.

Alternatively, consider that the coefficient of $x^2$ is positive, so the graph of $(x+1)(x+3)=0$ opens up. When there are two distinct roots, the shape of the parabola means that the product is negative when $x$ is between the roots and positive when $x$ is less than both roots or greater than both roots.