#### Answer

$x_{1} = -2 + \sqrt{3}i$ and $x_{2}= -2 -\sqrt{3}i$

#### Work Step by Step

Given $x^2+4x=-7 \longrightarrow x^2+4x+7=0 \\ $
$a= 1, \ b=4, \ c=7$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-4 \pm \sqrt{4^2-4 \times 1 \times 7}}{2 \times 1} = \dfrac{-4 \pm \sqrt{16-28}}{2} = \dfrac{-4 \pm \sqrt{-12}}{2} = \dfrac{-4 \pm \sqrt{4 \times (-3)}}{2} = \dfrac{-4 \pm 2\sqrt{ (-3)}}{2} = -2 \pm \sqrt{-3} = -2 \pm \sqrt{3}i$
Therefore the solutions are $x_{1} = -2 + \sqrt{3}i$ and $x_{2}= -2 -\sqrt{3}i$