Answer
$\left\{-\dfrac{\sqrt{b^2+16}}{2},\dfrac{\sqrt{b^2+16}}{2}\right\}$
Work Step by Step
Using the properties of equality, the given equation, $
4x^2=b^2+16
,$ is equivalent to
\begin{align*}\require{cancel}
\dfrac{\cancel4x^2}{\cancel4}&=\dfrac{b^2+16}{4}
\\\\
x^2&=\dfrac{b^2+16}{4}
.\end{align*}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{align*}
x&=\pm\sqrt{\dfrac{b^2+16}{4}}
\\\\
x&=\pm\sqrt{\left(\dfrac{1}{4}\right)(b^2+16)}
\\\\
x&=\pm\sqrt{\left(\dfrac{1}{2}\right)^2(b^2+16)}
\\\\
x&=\pm\dfrac{1}{2}\sqrt{b^2+16}
\\\\
x&=\pm\dfrac{\sqrt{b^2+16}}{2}
.\end{align*}
Hence, the solution set of the equation $
4x^2=b^2+16
$ is $\left\{-\dfrac{\sqrt{b^2+16}}{2},\dfrac{\sqrt{b^2+16}}{2}\right\}$.