Let's solve the equation \(25^{x} + 3(5^{x}) = 4\) for x. We can rewrite the equation as \((5^{2})^{x} + 3(5^{x}) = 4\), which simplifies to \(5^{2x} + 3(5^{x}) = 4\). Let's make a substitution: let \(y = 5^{x}\). 

Substituting this into our equation, we get \(y^2 + 3y = 4\). Rearranging, we have a quadratic equation: \(y^2 + 3y - 4 = 0\). 

Solving this quadratic equation using the quadratic formula, we find that the solutions for y are \(y = -4\) or \(y = 1\).

Recall that we made the substitution \(y = 5^{x}\). So, we have two equations to solve for x: \(5^{x} = -4\) and \(5^{x} = 1\). 

The first equation has no solution, since an exponential function with a positive base always returns a positive value. 

The second equation has one solution: \(x = 0\), since any non-zero number raised to the power of zero is equal to one.

So, the only solution to the original equation is x = 0.