#### Answer

$(3,-4)$

#### Work Step by Step

Completing the square of the right side of the given equation, $
y=x^2-6x+5
,$ results to
\begin{array}{l}\require{cancel}
y=(x^2-6x)+5
\\\\
y=\left( x^2-6x+\left( \dfrac{-6}{2} \right)^2 \right)+5-\left( \dfrac{-6}{2} \right)^2
\\\\
y=\left( x^2-6x+\left( -3 \right)^2 \right)+5-\left( -3 \right)^2
\\\\
y=\left( x^2-6x+9 \right)+5-9
\\\\
y=\left( x-3 \right)^2-4
.\end{array}
The vertex of the parabola whose quadratic equation is in the form $y=a(x-h)^2+k$ is $(h,k)$. Hence, the vertex of the equation above is
\begin{array}{l}\require{cancel}
(3,-4)
.\end{array}