The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?
Expanding the quadratic on the left gives
\[x^2-mx+14=0.\]The discriminant is $\sqrt{m^2-56}.$ To have real solutions for $x$, the discriminant must be nonnegative, so $m^2-56\ge 0$. Applying the difference of squares, $(m - 8)(m + 8) \ge 0$, so $m \le -8$ or $m \ge 8$. If $m = -8k$ with $k = 0,$ $1$, $2,$ or $3$, then $m^2-56 = 64k^2-56$ is positive. If $m = 8k$ with $k = 1$, $2, \dots$, $7$, then $m^2-56 = 64k^2-56$ is again positive. Thus, the answer is $3 + 7 = \boxed{10}$.
To find the values of $m$ for which the quadratic equation has integer roots, we need to use the discriminant of the quadratic equation.
The discriminant, denoted as $\Delta$, is given by the formula $\Delta = b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation in the form $ax^2 + bx + c = 0$.
In this case, the quadratic equation is $x^2-mx+24=10$. Comparing this equation with the general form, we have $a=1$, $b=-m$, and $c=24-10=14$.
Using the discriminant, we have $\Delta = (-m)^2 - 4(1)(14) = m^2 - 56$.
For the quadratic equation to have integer roots, the discriminant must be a perfect square. Therefore, $m^2 - 56 = k^2$, where $k$ is an integer.
Rearranging the equation, we have $m^2 - k^2 = 56$.
This equation is a difference of squares, which can be factored as $(m + k)(m - k) = 56$.
To find the possible values of $m$, we need to consider all possible factorizations of 56:
1) $m + k = 56$ and $m - k = 1$
Solving this system of equations, we get $m = \frac{57}{2}$, which is not an integer.
2) $m + k = 28$ and $m - k = 2$
Solving this system of equations, we get $m = 15$.
3) $m + k = 14$ and $m - k = 4$
Solving this system of equations, we get $m = 9$.
4) $m + k = 8$ and $m - k = 7$
Solving this system of equations, we get $m = 7.5$, which is not an integer.
Therefore, there are two different values of $m$ for which the quadratic equation has integer roots.
Answer: $\boxed{2}$ different values of $m$ are possible.
To find the possible values of $m$, we need to first solve the given quadratic equation $x^2-mx+24 = 10$.
Step 1: Subtract 10 from both sides of the equation to move the constant term to the right side:
\[x^2 - mx + 14 = 0.\]
Step 2: We can either use the quadratic formula or factorization to solve this equation. Let's use factorization here.
To factorize $x^2 - mx + 14$, we need to find two numbers whose product is $14$ and sum is $m$. Let's list the factors of $14$: $\pm 1, \pm 2, \pm 7, \pm 14$. We're looking for two integers, which means we can discard the negative factors.
Now, let's find the possible values of $m$ for each factor pair:
1) $1 \cdot 14$: In this case, $m = 1 + 14 = 15$.
2) $2 \cdot 7$: In this case, $m = 2 + 7 = 9$.
Step 3: Now that we have found the possible values of $m$ from the factor pairs, we need to determine how many of these result in integer roots for $x_1$ and $x_2$.
Considering the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ($ax^2 + bx + c = 0$), we know that for the roots to be integers, the discriminant $b^2 - 4ac$ must be a perfect square.
In our equation $x^2 - mx + 14 = 0$, we have $a = 1$, $b = -m$, and $c = 14$. The discriminant is: $(-m)^2 - 4(1)(14) = m^2 - 56$.
For $m = 15$, the discriminant becomes $15^2 - 56 = 169$, which is a perfect square ($13^2$). This means that $m = 15$ is a valid value.
For $m = 9$, the discriminant becomes $9^2 - 56 = 25$, which is a perfect square ($5^2$). This means that $m = 9$ is also a valid value.
Step 4: Conclusion. Since there are two valid values of $m$, namely $15$ and $9$, the final answer is that there are $\boxed{2}$ different values of $m$ possible for which $x_1$ and $x_2$ are integers.