The given quadratic equation is:

$h - m x^2 - 10x + 21 = 0$

We are told that $x = h$ and $x = m$ are the solutions to this equation, and $h > m$. Using the property of quadratic equations, the standard form can be written as:

$a x^2 + bx + c = 0$

where the solutions are $h$ and $m$. We can use Vieta's formulas to relate the coefficients of the quadratic equation to the roots $h$ and $m$:

1. The sum of the roots (from Vieta’s relations) is:

$h + m = -\frac{b}{a}$

2. The product of the roots is:

$h \cdot m = \frac{c}{a}$

Comparing the given quadratic equation with the standard form, we have:

$h - m = 1, \quad b = -10, \quad c = 21$

Therefore:

1. From the sum of the roots: $h + m = -\frac{-10}{1} = 10$.
2. From the product of the roots: $h \cdot m = \frac{21}{1} = 21$.

Now, we have the system of equations:

$h + m = 10 \quad \text{(1)}$ $h \cdot m = 21 \quad \text{(2)}$

To solve this system, substitute $h = 10 - m$ from equation (1) into equation (2):

$(10 - m) \cdot m = 21$

Expanding this:

$10m - m^2 = 21$

Rearrange into a standard quadratic form:

$m^2 - 10m + 21 = 0$

Now, solve for $m$ using the quadratic formula:

$m = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(21)}}{2(1)}$ $m = \frac{10 \pm \sqrt{100 - 84}}{2}$ $m = \frac{10 \pm \sqrt{16}}{2}$ $m = \frac{10 \pm 4}{2}$

Thus, we have two solutions:

$m = \frac{10 + 4}{2} = 7 \quad \text{or} \quad m = \frac{10 - 4}{2} = 3$

Since $h > m$, we choose $m = 3$, which gives $h = 10 - 3 = 7$.

Finally, the value of $h - m = 7 - 3 = 4$.

So, the value of $h - m$ is $4$.

Do you have any questions or need more details?

Here are five related questions to further explore this topic:

1. What are Vieta's formulas and how do they apply to quadratic equations?
2. How can the quadratic formula be derived?
3. What are the steps to solve a quadratic equation by factoring?
4. How does the discriminant determine the nature of the roots in a quadratic equation?
5. What are real-life applications of quadratic equations?

Tip: When solving quadratic equations, always check if factoring is possible before using the quadratic formula for simplicity.