Solve (x+2)^2+8=0

$$( x + 2 ) ^ { 2 } + 8 = 0$$

Solve for x (complex solution)

$$x=-2\sqrt{2}i-2\approx -2-2.828427125i$$

$$x=-2+2\sqrt{2}i\approx -2+2.828427125i$$

Steps by Finding Square Root

Subtracting 8 from itself leaves 0.

$$\left(x+2\right)^{2}=-8$$

Take the square root of both sides of the equation.

$$x+2=2\sqrt{2}i x+2=-2\sqrt{2}i$$

Subtract 2 from both sides of the equation.

$$x+2-2=2\sqrt{2}i-2 x+2-2=-2\sqrt{2}i-2$$

Subtracting 2 from itself leaves 0.

$$x=2\sqrt{2}i-2 x=-2\sqrt{2}i-2$$

Subtract 2 from 2i\sqrt{2}.

$$x=-2+2\sqrt{2}i$$

Subtract 2 from -2i\sqrt{2}.

$$x=-2\sqrt{2}i-2$$

The equation is now solved.

$$x=-2+2\sqrt{2}i x=-2\sqrt{2}i-2$$