Answer :

Let's solve the equation [tex]\(\sqrt{3x + 3} - 1 = x\)[/tex].

1. Step 1: Isolate the square root term.
[tex]\[ \sqrt{3x + 3} - 1 = x \][/tex]
Add 1 to both sides:
[tex]\[ \sqrt{3x + 3} = x + 1 \][/tex]

2. Step 2: Square both sides to eliminate the square root.
[tex]\[ (\sqrt{3x + 3})^2 = (x + 1)^2 \][/tex]
Simplify:
[tex]\[ 3x + 3 = (x + 1)^2 \][/tex]
[tex]\[ 3x + 3 = x^2 + 2x + 1 \][/tex]

3. Step 3: Rearrange the equation to set it to zero.
[tex]\[ 3x + 3 = x^2 + 2x + 1 \][/tex]
Subtract [tex]\(3x + 3\)[/tex] from both sides:
[tex]\[ 0 = x^2 + 2x + 1 - 3x - 3 \][/tex]
[tex]\[ 0 = x^2 - x - 2 \][/tex]

4. Step 4: Solve the quadratic equation.
The quadratic equation is:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 2)(x + 1) = 0 \][/tex]
Set each factor to zero:
[tex]\[ x - 2 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = 2 \quad \text{or} \quad x = -1 \][/tex]

5. Step 5: Verify the solutions.
Substitute [tex]\(x = 2\)[/tex] back into the original equation:
[tex]\[ \sqrt{3(2) + 3} - 1 = 2 \][/tex]
[tex]\[ \sqrt{6 + 3} - 1 = 2 \][/tex]
[tex]\[ \sqrt{9} - 1 = 2 \][/tex]
[tex]\[ 3 - 1 = 2 \][/tex]
[tex]\[ 2 = 2 \quad \text{(True)} \][/tex]

Substitute [tex]\(x = -1\)[/tex] back into the original equation:
[tex]\[ \sqrt{3(-1) + 3} - 1 = -1 \][/tex]
[tex]\[ \sqrt{-3 + 3} - 1 = -1 \][/tex]
[tex]\[ \sqrt{0} - 1 = -1 \][/tex]
[tex]\[ 0 - 1 = -1 \][/tex]
[tex]\[ -1 = -1 \quad \text{(True)} \][/tex]

Both solutions satisfy the original equation.

Therefore, the correct answer is:
A. [tex]\(-1\)[/tex] and [tex]\(2\)[/tex]