What are all roots of the equation (0.6^x)*(25/9)^(x^2-12)=(27/125)^3

We'll write the 1st factor as a
fraction:

0.6^x = (6/10)^x =
(3/5)^x

We notice that 25/9 =
(5/3)^2

We'll raise both sides by (x^2 -
12):

(25/9)^(x^2 - 12)  = (5/3)^2*(x^2 -
12)

We notice that 27/125 =
(3/5)^3

We'll raise both sides by
3:

(27/125)^3 = (3/5)^9

We'll
write the equation:

[(3/5)^x]*[(5/3)^2*(x^2 - 12)] =
(3/5)^9

But [(5/3)^2*(x^2 - 12)] = [(3/5)^-2*(x^2 -
12)]

Since the bases form the left side are matching, we'll
add the exponents:

[(3/5)^(x-2*(x^2 - 12))] =
(3/5)^9

Since the bases form the left side are matching,
we'll apply one to one rule:

x-2*(x^2 - 12) =
9

We'll remove the brackets:

x
- 2x^2 + 24 - 9 = 0

-2x^2 + x + 15 =
0

We'll calculate the roots of the
equation:

x1 =
[-1+sqrt(1+120)]/-4

x1 =
(-1+11)/-4

x1 = -5/2

x2 =
3

The required roots of the equation are:
{-5/2 ; 3}.