1+5^x -(2*25*x)=0

Let us simplify the function:

1+ 5^x - 2*5^2x)=0

Now assume that y=5^x

1+y-2y^2 =0

Now we have a function is the standard formula :

2y^2 -y-1=0

Factorize:

(2y+1)(y-1)=0

y1= -1/2 ==> y1=5^x1 = -1/2 which is impossible

y2= 1 ==> y2= 5^x2 = 1 ==> x2=0

The the solution for the function is x=0

1+5^x-2*25^x = 0 is a quadratic in 5^x.Or

1+y -2y^2 = 0. Where y = 5^x.

(-1) (2y^2-y-1) = 0.

(-1)(2y+1)(y-1) = 0. Or

2y+1 = 0. Or y - 1 = 0.

2y+1= 0 gives y = -1/2. Or 5^x = (-1/2) which has no solution in R

y -1 = 0 gives y=1 or 5^x =1 = 5^(0). Or x = 0, the real solution

We'll substitute 25 by 5^2 so we can re-write the eq. in this manner:

-2*(5^2)^x + 5^x + 1=0

We'll substitute 5^x by t.

-2*t^2 + t +1=0

We'll apply the quadratic formula:

t1=[-1+sqrt (1+4*2)]/-4

t1=(-1+3)/-4

t1=2/-4

We'll divide by 2:

**t1=-1/2**

t2=(-1-3)/-4

t2=-4/-4

**t2=1**

**Now we'll have to find out x values:**

5^x=t1

5^x=-1/2 impossible, because 5^x>0!

5^x=t2=1

5^x = 5^0

x=0

The only solution of the equation is x=0.