Answer
$r=\frac{3}{2}+i \frac{3 \sqrt{3}}{2}$
$r = \frac{3}{2}-i \frac{3 \sqrt{3}}{2}$
Work Step by Step
Given \begin{equation}
4 r^2-12 r+36=0.
\end{equation} This equation can be best solved by the quadratic formula. First factor out $4$:
\begin{equation}
\begin{aligned}
4(r^2-3 r+9)&=0 \\
r^2-3 r+9&= 0.
\end{aligned}
\end{equation} Apply the quadratic formula:
\begin{equation}
\begin{aligned}
a & =1, b=-3, c=9 \\
r & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
r & =\frac{-(-3) \pm \sqrt{(-3)^2-4 \cdot 1 \cdot 9}}{2 \cdot 1}\\
&= \frac{3\pm \sqrt{-27}}{2}\\
&= \frac{3\pm i\sqrt{9\cdot 3}}{2}\\
& =\frac{3\pm i3\sqrt{ 3}}{2}\\
\Longrightarrow r & =\frac{3}{2}+i \frac{3 \sqrt{3}}{2} \\
r & =\frac{3}{2}-i \frac{3 \sqrt{3}}{2}.
\end{aligned}
\end{equation} Check \begin{equation}
\begin{aligned}
\left( \frac{3}{2}+i \frac{3 \sqrt{3}}{2} \right)^2-3\left( \cdot \frac{3}{2}+i \frac{3 \sqrt{3}}{2}\right)+9& \stackrel{?}{=} 0 \\
0 & =0 \checkmark \\
\left( \frac{3}{2}-i \frac{3 \sqrt{3}}{2} \right)^2-3\left( \cdot \frac{3}{2}-i \frac{3 \sqrt{3}}{2}\right)+9& \stackrel{?}{=} 0 \\
0 & =0 \checkmark.
\end{aligned}
\end{equation} The solution is \begin{equation}
\begin{aligned}
r&=\frac{3}{2}+i \frac{3 \sqrt{3}}{2}\\
r& = \frac{3}{2}-i \frac{3 \sqrt{3}}{2}.
\end{aligned}
\end{equation}
