Given

$\rm 5^{x - 2} + 5^{3 - x} = 6$

By using the law of indices, $\rm a^{m-n} = \frac{a^{m}}{a^{n}}$,

$\rm or, \frac{5^{x}}{5^{2}} + \frac{5^{3}}{5^{x}} = 6$

$\rm or, \frac{5^{x}}{25} + \frac{125}{5^{x}} = 6$

Let $\rm 5^{x} = a$

$\rm or, \frac{a}{25} + \frac{125}{a} = 6$

Multiplying both sides of the equation by 25a, we get,

$\rm or, 25a \cdot \frac{a}{25} + 25a \cdot \frac{125}{a} = 25a \cdot 6$

$\rm or, a^{2} + 3125 = 150a$

$\rm or, a^{2} - 150 a + 3125 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve it.

$\rm or, a^{2} - (125 + 25) a + 3125 = 0$

$\rm or, a^{2} - 125 a - 25 a + 3125 = 0$

$\rm or, a (a - 125) - 25 (a - 125) = 0$

$\rm or, (a - 25) (a - 125) = 0$

Either

$\rm (a - 25) = 0$

$\rm or, a = 25$

$\rm or, a = 5^{2}$

By supposition, $\rm a = 5^{x}$. We substitute the value for a and we get,

$\rm or, 5^{x} = 5^{2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 2$

Or

$\rm (a - 125) = 0$

$\rm or, a = 125$

$\rm or, a = 5^{3}$

By supposition, $\rm a = 5^{x}$. We substitute the value for a and we get,

$\rm or, 5^{x} = 5^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required values of x are x = {2,3}.