Solve the Quadratic Inequality: When is x² - 6x + 10 Less Than Zero?

To solve for the values of $x$ such that $f(x) = x^2 - 6x + 10 < 0$, we first identify the vertex of the quadratic function.

1. Calculate the vertex $x$ coordinate:
The formula for the vertex's $x$-coordinate of a quadratic function $ax^2 + bx + c$ is $x = -\frac{b}{2a}$. For our function, $a = 1$ and $b = -6$, thus:

$x = -\frac{-6}{2 \times 1} = 3$

2. Calculate the minimum $y$ value at this $x$ (point of vertex):

Substitute $x = 3$ back into the function:

$y = 3^2 - 6 \cdot 3 + 10 = 9 - 18 + 10 = 1$

3. Since the parabola opens upwards (because $a = 1 > 0$) and its minimum value is 1 (greater than 0), the function does not achieve any negative values.

Therefore, the quadratic function $y = x^2 - 6x + 10$ does not have any negative values for any real number $x$.

The correct answer to the problem is: The function has no negative values.