Answer
$k=8+\sqrt{67}, k=8-\sqrt{67}$
Work Step by Step
We start by completing the square in line $1$ of the equation by dividing the coefficient of the $16k$ term by $2$ and squaring it and adding the result on both sides.
$$
\begin{aligned}
k^2-16k & = 3 \\
k^2-16k+\left(\frac{16}{2}\right)^2 & =3+\left(\frac{16}{2}\right)^2 \\
k^2-16k+8^2 & =3+8^2 \\
(k-8)^2 & =67.
\end{aligned}
$$ Take the square root:
$$
\begin{array}{rl}
(k-8)^2 & =67 \\
k-8 & = \pm \sqrt{67} \\
k-8= \sqrt{67} &\ \textbf{or}\ \ \ k-8=-\sqrt{67}\\
k=8+\sqrt{67} & \ \textbf{or}\ \ \ k= 8-\sqrt{67}.\end{array}
$$ Check $$\begin{aligned}
(8+\sqrt{67})^2-16(8+\sqrt{67}) & \stackrel{?}{=} 3\\
64+16\sqrt{67}+67-128-16\sqrt{67}& \stackrel{?}{=}3\\
3& = 3\checkmark.
\end{aligned}$$ $$\begin{aligned}
(8-\sqrt{67})^2-16(8-\sqrt{67}) & \stackrel{?}{=} 3\\
64-16\sqrt{67}+67-128+16\sqrt{67}& \stackrel{?}{=}3\\
3& = 3\checkmark.
\end{aligned}$$ The solution is $$k=8+\sqrt{67}, k=8-\sqrt{67}.$$
