The given equation is \(125_x = 20_{10}\), where 125 is a number in base `x` and 20 is a number in base 10.

To find the value of `x`, we can convert both numbers to the same base and then solve for `x`.

Let's convert the number 125 in base `x` to base 10. We can do this by multiplying each digit by its place value and adding them up.

The place value of a digit in a number system is determined by the base of the number system raised to the power of the position of the digit from right to left, starting from 0.

So, for the given number 125 in base `x`, we have:

\(1 \times x^2 + 2 \times x^1 + 5 \times x^0 = x^2 + 2x + 5\)

Now, we can set this expression equal to 20 and solve for `x`:

\(x^2 + 2x + 5 = 20\)

\(x^2 + 2x - 15 = 0\)

This is a quadratic equation, which we can solve using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting the values of `a`, `b`, and `c` into this formula, we get:

\(x = \frac{-(2) \pm \sqrt{(2)^2 - 4(1)(-15)}}{2(1)}\)

\(x = \frac{-2 \pm \sqrt{64}}{2}\)

\(x = (-2 \pm 8) / 2\)

So, there are two possible values for `x`: \((-2 + 8) / 2 = 3\) and \((-2 - 8) / 2 = -5\). However, since a base must be a positive integer, only 3 is a valid solution. So, the value of `x` that satisfies the given equation is 3,