1x^2 - 8x + 16 = 0

To solve the quadratic equation \(1x^2 - 8x + 16 = 0\) using the quadratic formula, we can identify the coefficients:

- \(a = 1\)
- \(b = -8\)
- \(c = 16\)

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

First, calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 = (-8)^2 = 64
\]
\[
4ac = 4 \cdot 1 \cdot 16 = 64
\]
\[
b^2 - 4ac = 64 - 64 = 0
\]

Since the discriminant is 0, there will be one real solution (a repeated root):

Now substitute the values into the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1}
\]
\[
x = \frac{8 \pm 0}{2}
\]
\[
x = \frac{8}{2} = 4
\]

Thus, the equation \(1x^2 - 8x + 16 = 0\) has one repeated solution:

\[
\boxed{4}
\]