Answer
$(-\infty,-2-\sqrt 3)U(-2+\sqrt 3,\infty)$
Work Step by Step
Step 1. Rewrite the inequality as $x^2+4x+1\gt0$
Step 2. Solve the equation $x^2+4x+1=0$ to get $x_1=-2-\sqrt 3, x_2=-2+\sqrt 3$ which are the boundary points.
Step 3. Use test points $x=-4, -1, 0$, the signs of the left side quadratic are $+, -, +$
Step 4. The inequality requires positive (two outer regions), thus the solution intervals are $(-\infty,-2-\sqrt 3)U(-2+\sqrt 3,\infty)$
