How about this one: Solve 5^(2x)+ 5^(x+1) - 6 =0

I did the math already i just want to confirm my answers.
t=1 t= -6
x=0 x=log(-6)/log(5)
is this right?

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5^(2x)+ 5^(x+1) - 6 = 0            ← this can be written as a quadratic equation in 5ˣ

     (5ˣ)² + 5(5ˣ) - 6 = 0            ← this is like u²+5u-6=0 for u=5ˣ
      (5ˣ + 6)(5ˣ - 1) = 0            ← this is like (u+6)(u-1) = 0
Possible solutions are for:
➊ 5ˣ + 6 = 0   ⇒   5ˣ = -6                       ← There is no solution to this
                                                                    since 5 to any power cannot be negative
➋ 5ˣ - 1 = 0   ⇒   5ˣ = 1   ⇒   x = 0   ✔

           ANSWER
               x = 0


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I'm not sure how you got your answers. The way to do this is to substitute u = 5^x; u^2 = 5^2x:

5^2x + 5 (5^x) - 6 = 0
u^2 + 5u - 6 = 0
(u-1)(u+6) = 0
u = 1 or -6, as you noted.
5^x = 1 or -6

x = 0 is the only real solution.

Great, except the t=-6 leads to complex numbers. I assume you should call it a non-answer!

i.e., log(-6) fails on most calculators!

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

(5^x + 6)(5^x - 1) = 0

5^x = -6 (not possible)

5^x = 1 --> x = 0

CORRECT!