Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book. I'm studying on how to find the vertex of a Parabola and got stuck with this question.

Determine the vertex of $y = x^2 - 6x + 1$

I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it does not have a section where it shows how it got to that answer so I'm doubting if my understanding is incorrect. If I use online algebra calculators, it matches with mine but I just want to make sure that I understand how signs and the formula work.

Formula I used to look for the Vertex:
$$V=\left(-\frac{b}{2a} , \frac{4ac-b^2}{4a}\right)$$
My solution:
$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{-32}{4} = \frac{-8}{1}=-8
\end{align}$$
$$V=(3,-8)$$
What I think was done on the book:
$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{40}{4} = \frac{10}{1}=10
\end{align}$$
$$V=(3,10)$$
Is it correct that I should've added $4-36$ since they are both positive numbers or subtract it?
Any explanation is appreciated. Thanks!

An easier way to do this is to take a derivative, solve for the minimum, this will give you, your x-value. The substituting that value back into the original equation we can get the y-value. For example here you have:
y=(x^2)-6x+1
derivative: y'=(2x-6)
Solving for x-value (solving for when y'=0) : 2x-6=0 -> x=3   <-X-value
Now that we have the x value we can use this value in our original equation
y=3^2 -6*3 +1 We find that y=-8
This gives us the coordinates for the vertex. This is probably the fastest approach here as taking the derivative of this function is pretty basic.. Hopefully this helps.
