Comparing $\rm x^{2} - 9x + 20 = 0$ with the general quadratic equation $\rm ax^{2} + bx + c = 0$, we get,

$\rm a = 1, b = -9, c = 20$

Using the quadratic formula to find the roots of the above quadratic equation:

$\rm x = \frac{ - b \pm \sqrt{ b^{2} - 4ac }}{2a}$

First, we calculate the discriminant of the quadratic equation:

$\rm (b^{2} - 4ac) = (-9)^{2} - 4 \cdot 1 \cdot 20 = 81 - 80 = 1$

Now,

$\rm x = \frac{ - b \pm \sqrt{ b^{2} - 4ac}}{2a}$

$\rm or, x = \frac{ - (-9) \pm \sqrt{1}}{2 \cdot 1}$

$\rm or, x = \frac{ 9 \pm 1}{2}$

Taking positive sign, we get,

$\rm x = \frac{ 9 + 1}{2}$

$\rm or, x = \frac{10}{2}$

$\rm \therefore x = 5$

Taking negative sign, we get,

$\rm x = \frac{ 9 - 1}{2}$

$\rm or, x = \frac{8}{2}$

$\rm \therefore x = 4$

Hence, the required values of x are x = {4, 5}.