Solve 2(z-2b)^2=a

$$2 ( z - 2 b ) ^ { 2 } = a$$

Solve for a

$$a=2\left(z-2b\right)^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(z-2b\right)^{2}.

$$2\left(z^{2}-4zb+4b^{2}\right)=a$$

Use the distributive property to multiply 2 by z^{2}-4zb+4b^{2}.

$$2z^{2}-8zb+8b^{2}=a$$

Swap sides so that all variable terms are on the left hand side.

$$a=2z^{2}-8zb+8b^{2}$$

Solve for b (complex solution)

$$b=\frac{z}{2}+\frac{\sqrt{2a}}{4}$$

$$b=\frac{z}{2}-\frac{\sqrt{2a}}{4}$$

Solve for b

$$b=\frac{z}{2}+\frac{\sqrt{2a}}{4}$$

$$b=\frac{z}{2}-\frac{\sqrt{2a}}{4}\text{, }a\geq 0$$