Therefore, the solutions for y are y = -8/5 and y = -3.
Now we can substitute back y = 5^x:
5^x = -8/5 and 5^x = -3
These equations do not have real solutions because 5 raised to any power will always be positive. Therefore, the original equation 5(5^(2x)) + 43(5^x) + 24 = 0 has no real solutions.
Let's simplify the equation step by step:
5(5^(2x)) + 43(5^x) + 24 = 0
Let's substitute y = 5^x. Then we have:
5(y^2) + 43(y) + 24 = 0
Now we have a quadratic equation in terms of y:
5y^2 + 43y + 24 = 0
Let's factor this quadratic equation:
(5y + 8)(y + 3) = 0
Therefore, the solutions for y are y = -8/5 and y = -3.
Now we can substitute back y = 5^x:
5^x = -8/5 and 5^x = -3
These equations do not have real solutions because 5 raised to any power will always be positive. Therefore, the original equation 5(5^(2x)) + 43(5^x) + 24 = 0 has no real solutions.