Solve (2x+4b^0)^2+8b^0=180^0

$$( 2 x + 4 b ^ { 0 } ) ^ { 2 } + 8 b ^ { 0 } = 180 ^ { 0 }$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{7}i}{2}-2$$

$$x=\frac{\sqrt{7}i}{2}-2\text{, }b\neq 0$$

Steps by Finding Square Root

Subtracting 8 from itself leaves 0.

$$\left(2x+4\right)^{2}=1-8$$

Subtract 8 from 1.

$$\left(2x+4\right)^{2}=-7$$

Take the square root of both sides of the equation.

$$2x+4=\sqrt{7}i 2x+4=-\sqrt{7}i$$

Subtract 4 from both sides of the equation.

$$2x+4-4=\sqrt{7}i-4 2x+4-4=-\sqrt{7}i-4$$

Subtracting 4 from itself leaves 0.

$$2x=\sqrt{7}i-4 2x=-\sqrt{7}i-4$$

Subtract 4 from i\sqrt{7}.

$$2x=-4+\sqrt{7}i$$

Subtract 4 from -i\sqrt{7}.

$$2x=-\sqrt{7}i-4$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{-4+\sqrt{7}i}{2} \frac{2x}{2}=\frac{-\sqrt{7}i-4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{-4+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-4}{2}$$

Divide i\sqrt{7}-4 by 2.

$$x=\frac{\sqrt{7}i}{2}-2$$

Divide -i\sqrt{7}-4 by 2.

$$x=-\frac{\sqrt{7}i}{2}-2$$

The equation is now solved.

$$x=\frac{\sqrt{7}i}{2}-2 x=-\frac{\sqrt{7}i}{2}-2$$

Solve for b (complex solution)

$$b\neq 0$$

$$x=-\frac{\sqrt{7}i}{2}-2\text{ or }x=\frac{\sqrt{7}i}{2}-2$$