Question

$${ a }^{ 2 } = { b }^{ 2 } + { d }^{ 2 }$$

Solve for b (complex solution)

$b=-\sqrt{a^{2}-d^{2}}$
$b=\sqrt{a^{2}-d^{2}}$

Solve for a

$a=\sqrt{b^{2}+d^{2}}$
$a=-\sqrt{b^{2}+d^{2}}$

Solve for b

$b=\sqrt{a^{2}-d^{2}}$
$b=-\sqrt{a^{2}-d^{2}}\text{, }|a|\geq |d|$