You need to come up with the following varaible replacement, such that:

`5^x = y => y^2 - 6y + 5 = 0`

You may use factorization to solve quadratic equation, such that:

`y^2 - (5 + 1)y + 5 = 0 => y^2 - 5y - y + 5 = 0`

`(y^2 - 5y) - (y - 5) = 0 => y(y - 5) - (y - 5) = 0`

Factoring out (y - 5) yields:

`(y - 5)(y - 1) = 0`

Using the zero product rule yields:

`y - 5 = 0 => y = 5`

`y - 1 = 0 => y = 1`

Replacing back `5^x = y` yields:

`5^x = 5 => ln(5^x) = ln 5 => x*ln 5 = ln 5 => x = (ln 5)/(ln 5) => x = 1`

`5^x = 1 => x*ln 5 = ln 1 => x = (ln 1)/(ln 5) => x = (0)/(ln 5) => x = 0`

**Hence, evaluating the solutions to the given equation yields **`x = 0, x = 1.`

This is an exponential equtaion and we'll solve it using technique of substitution.

We'll substitute 5^x by the variable t.

25^x=(5^2)^x=(5^x)^2=t^2

We'll re-write the equation in t:

t^2 -6t + 5 = 0

We'll apply the quadratic formula:

t1 = [6+sqrt(36-20)]/2

t1 = (6+4)/2

t1 = 5

t2 = (6-4)/2

t2 = 1

Now, we'll determine x:

5^x = t1

5^x = 5

Since the bases are matching, we'll apply one to one property:

x = 1

5^x = t2

5^x = 1

But 5^0 = 1

5^x = 5^0

x = 0

The solutions of the equation are: {0 ; 1}.

The equation `25^x - 6*5^x + 5 = 0` has to be solved.

`25^x - 6*5^x + 5 = 0`

`(5^2)^x - 6*5^x + 5 = 0`

`(5^x)^2 - 6*5^x + 5 = 0`

`(5^x)^2 - 5*5^x - 5^x + 5 = 0`

`5^x(5^x - 5) -1(5^x - 5) = 0`

`(5^x - 1)(5^x - 5) = 0`

`5^x - 1 = 0`

`5^x = 1`

x = 0

`5^x - 5 = 0`

`5^x = 5`

x = 1

The solution of the equation `25^x - 6*5^x + 5 = 0` is x = 0 and x = 1