Answer
The solution set is $\left\{1, 7\right\}$.
Work Step by Step
Subtract $10$ from both sides: to obtain:
$$2(k-4)^2=18$$
Divide $2$ to both sides:
\begin{align*}
\frac{2(x-4)^2}{2}&=\frac{18}{2}\\\\
(x-4)^2&=9
\end{align*}
Take the square root of both sides to obtain:
\begin{align*}
\sqrt{(x-4)^2}&=\pm\sqrt9\\
x-4&=\pm3\\
x&=4\pm3\\
x_1&=4-3=1\\
x_2&=4+3=7
\end{align*}
Checking:
\begin{align*}
2(1-4)^2+10&=28\\
2(-3)^2+10&=28\\
2(9)+10&=28\\
18+10&=28\\
28&\stackrel{\checkmark}=28
\end{align*}
\begin{align*}
2(7-4)^2+10&=28\\
2(3)^2+10&=28\\
2(9)+10&=28\\
18+10&=28\\
28&\stackrel{\checkmark}=28
\end{align*}
Thus, the solution set is $\left\{1, 7\right\}$.