#### Answer

$\left\{2, 8\right\}$

#### Work Step by Step

Subtract $10x$ to both sides:
$x^2+16-10x=10x-10x
\\x^2-10x+16=0$
RECALL:
A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$.
The trinomial's factored form will be:
$x^2+bx+c=(x+d)(x+e)$
The trinomial above has $b=-10$ and $c=16$.
Note that $16=-8(-2)$ and $-10= -8+(-2)$.
This means that $d=-8$ and $e=-2$
Thus, the factored form of the trinomial is: $[x+(-8)][x+(-2)] = (x-8)(x-2)$
The given equation may be written as:
$(x-8)(x-2)=0$
Use the Zero-Product Property by equating each factor to zero.
Then, solve each equation to obtain:
$\begin{array}{ccc}
&x-8= 0 &\text{ or } &x-2=0
\\&x=8 &\text{ or } &x=2
\end{array}$
Thus, the solution set is $\left\{2, 8\right\}$.