You need to use the property of logarithmic function, such that:

`log_3 (x^2 + 4x + 12) = 2 => log_3 (x^2 + 4x + 12) = 2log_3 3`

`log_3 (x^2 + 4x + 12) = log_3 3^2 => log_3 (x^2 + 4x + 12)

= `log_3 9 =gt x^2 + 4x + 12 = 9 =gt x^2 + 4x = -3` `

You need to complete the square `x^2 + 4x` using the formula `(a + b)^2 = a^2 + 2ab + b^2` , such that:

`x^2 + 4x + 4 = 4 - 3 => (x + 2)^2 = 1 => x + 2 = +-1`

`x_1 = -2 + 1 => x_1 = -1`

`x_2 = -2 - 1 => x_2 = -3`

You need to test the values `x = -1` and `x = -3` in logarithmic equation such that:

`log_3 ((-1)^2 + 4(-1) + 12) = 2 => log_3 (1 - 4 + 12) = 2

`=>log_3 9 = 2 =gt log_3 3^2 = 2 =gt 2 log_3 3 = 2 =gt 2 = 2`

`log_3 ((-3)^2 + 4(-3) + 12) = 2 => log_3 (9 -12 + 12) = 2

=> `log_3 9 = 2 =gt 2 = 2` `

**Hence, testing both values `x = -1` and `x = -3` yields that the logarithmic equation holds.**

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}.