Answer
$x_{1}=-1$
$x_{2}=-3$
Work Step by Step
We have to solve: $2^{x^2+4x}=\frac{1}{8}$. we will use the principle of exponential equality.
$$\dfrac{1}{8}=\dfrac{1}{2^3}=2^{-3}.$$
So
$$2^{x^2+4x}=2^{-3}$$
or
$$x^2+4x=-3$$
$$x^2+4x+3=0$$
It's just a quadratic equation. $D=b^2-4ac$. In our case
$$D=4^2-4\cdot3=16-12=4>0$$
$$x_{1}=\dfrac{-4+\sqrt D}{2}=\dfrac{-4+2}{2}=-1$$
$$x_{2}=\dfrac{-4-\sqrt D}{2}=\dfrac{-4-2}{2}=-3$$
