Sure, let's solve the quadratic equation [tex]\(x^2 + 6x - 6 = 0\)[/tex] by completing the square step-by-step.
### Step 1: Move the constant term to the right side of the equation
[tex]\[ x^2 + 6x = 6 \][/tex]
### Step 2: Complete the square
To complete the square, we need to add and subtract a specific value on the left-hand side to make it a perfect square trinomial. The value to add and subtract is [tex]\(\left(\frac{b}{2}\right)^2\)[/tex], where [tex]\(b\)[/tex] is the coefficient of [tex]\(x\)[/tex].
Here, [tex]\(b = 6\)[/tex], so:
[tex]\[ \left(\frac{6}{2}\right)^2 = 3^2 = 9 \][/tex]
Add and subtract 9 on the left side:
[tex]\[ x^2 + 6x + 9 - 9 = 6 \][/tex]
Notice that [tex]\(x^2 + 6x + 9\)[/tex] is a perfect square trinomial:
[tex]\[ x^2 + 6x + 9 = (x + 3)^2 \][/tex]
Thus, we rewrite the equation as:
[tex]\[ (x + 3)^2 - 9 = 6 \][/tex]
### Step 3: Simplify and solve the equation for [tex]\(x\)[/tex]
Move [tex]\( -9 \)[/tex] to the right side:
[tex]\[ (x + 3)^2 = 6 + 9 \][/tex]
[tex]\[ (x + 3)^2 = 15 \][/tex]
Take the square root of both sides:
[tex]\[ x + 3 = \pm \sqrt{15} \][/tex]
Finally, solve for [tex]\(x\)[/tex] by subtracting 3 from both sides:
[tex]\[ x = -3 \pm \sqrt{15} \][/tex]
### Conclusion
The solutions to the equation [tex]\(x^2 + 6x - 6 = 0\)[/tex] are:
[tex]\[ x = -3 + \sqrt{15} \quad \text{or} \quad x = -3 - \sqrt{15} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{D. \, x=-3 \pm \sqrt{15}} \][/tex]
