## Elementary Algebra

Step 1: Comparing $x^{2}-16x+64=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=-16$ and $c=64$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $x=\frac{-(-16) \pm \sqrt {(-16)^{2}-4(1)(64)}}{2(1)}$ Step 4: $x=\frac{-16 \pm \sqrt {256-256}}{2}$ Step 5: $x=\frac{-16 \pm \sqrt {0}}{2}$ Step 6: $x=\frac{-16 \pm 0}{2}$ Step 7: $x=\frac{-16}{2}$ Step 8: $x=-8$ Step 9: Therefore, the solution set is {$-8$}. Therefore, the solution set contains only one integer which means that the question statement is false.