Solve equation 5^(2*x) = 5^(x)+5

First lets put all terms on the LHS;5^2x-5^x-5 = 0 (1);Now, to help us observe a pattern, put brackets on each term and separate power of x from its coefficients(5^x)²- (5^x) - 5 = 0 (2);Note that if we substitute u = 5^x we will start observing the pattern of a quadratic equation:u² - u -5 = 0 (3); Now, if you know how to solve a quadratic equation it is very straightforward - I like to use the method of finding the determinant first and the using it to finds roots using standard formula.The determinant: D = (-1)² - 4*(-5) = 21 (4);Using the standard formula for roots of a quadratic equation:u₁ = (-1 - sqrt(D))/2 = -1.791u₂ = (-1 + sqrt(D))/2 = 2.791Note that root of u₁ does not fit because it is impossible to obtain a negative number by raising a positive number by a real power. Therefore we can only use u₂ = 2.791; however, this is not our answer yet. To find the final answer we must equate u₂ = 5^x = 2.791 and take a logarithm of base 5. This gives us:log₅(5^x) = log₅(2.791) => x = 0.638The final answer is x = 0.638.QED