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!

$4-36$ is indeed $-32$. Whenever you have $a - b$ and $b > a$, you should interpret it as $a - b = -1 \cdot (b - a)$. So your calculation seems to be right. And yes, the vertex of the parabola is at $(3, -8)$, this can be verified by differentiating and setting $y' = 0$.
$y' = 2x - 6 = 0 \implies x = 3$ and then $y(x=3) = 3^2 - 6 \cdot 3 + 1 = 9 - 18 + 1 = -8$. Your method is correct, I just mentioned the differentiation method to confirm the answer.
