log3 (x^2+4x+12) = 2. To solve this we proceed as below:

We know if loga(b) = x, then b = a^x.

Therefore x^2+4x+12 = 3^2

x^2+4x+12 = 9

x^2+4x+12-9 = 0.

x^2+4x+3 = 0

We proceed to factorise the left:

x^2+3x+x+3 = 0

x(x+3) +1(x+3) = 0

(x+3)(x+1) = 0

Equate each factor to zero, as right side is zero.

x+3 = 0 or x+1 = 0

x +3 = 0 gives x = -3.

x + 1 = 0 gives x = -1.

Therefore x = -1, or x= -3 are the solution to the given logarithmatic equation.

First, we'll verify if the argument of the logarithm is positive. For this reason, we'll calculate the discriminant of the quadratic.

If the discriminant is negative and the coefficient of x^2 is positive, then the expression x^2 + 4x + 12 is positive for any value of x.

delta = b^2 - 4ac

We'll identify the coefficients a,b,c:

a = 1

b = 4

c = 12

delta = 16 - 4*12

delta = 16 - 48

delta = -32

Since delta is negative and a is positive, the expression x^2 + 4x + 12 > 0.

Now, we'll solve the equation. We'll take anti-logarithm:

x^2 + 4x + 12 = 3^2

x^2 + 4x + 12 = 9

We'll subtract 9 both sides:

x^2 + 4x + 12 - 9 = 0

We'll combine like terms:

x^2 + 4x + 3 = 0

We'll apply the quadratic formula:

x1 = [-4 +/- sqrt(16-12)]/2

x1 = (-4+2)/2

x1 = -1

x2 = -3

Since all values of x are admissible, we'll not reject either of resulted roots.

**The solutions of the equation are: {-3 ; -1}.**