Solve (4-x)^2+4=0

$$( 4 - x ) ^ { 2 } + 4 = 0$$

Solve for x (complex solution)

$$x=4+2i$$

$$x=4-2i$$

Steps by Finding Square Root

Subtracting 4 from itself leaves 0.

$$\left(-x+4\right)^{2}=-4$$

Take the square root of both sides of the equation.

$$-x+4=2i -x+4=-2i$$

Subtract 4 from both sides of the equation.

$$-x+4-4=2i-4 -x+4-4=-2i-4$$

Subtracting 4 from itself leaves 0.

$$-x=2i-4 -x=-2i-4$$

Subtract 4 from 2i.

$$-x=-4+2i$$

Subtract 4 from -2i.

$$-x=-4-2i$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{-4+2i}{-1} \frac{-x}{-1}=\frac{-4-2i}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{-4+2i}{-1} x=\frac{-4-2i}{-1}$$

Divide -4+2i by -1.

$$x=4-2i$$

Divide -4-2i by -1.

$$x=4+2i$$

The equation is now solved.

$$x=4-2i x=4+2i$$