It seems like there's a small typo in the equation you provided. The term `2x7` is unclear. I assume you meant to write `2x^2`. If that's correct, the equation should be:
\[ 2x^2 + 12x + 3 = 0 \]
Here's how you solve for \(x\) by completing the square:
1. **Divide the entire equation by the coefficient of \(x^2\) (which is 2 in this case):**
\[
\frac{2x^2 + 12x + 3}{2} = \frac{0}{2}
\]
Simplifying, we get:
\[
x^2 + 6x + \frac{3}{2} = 0
\]
2. **Move the constant term to the right side of the equation:**
\[
x^2 + 6x = -\frac{3}{2}
\]
3. **Complete the square:**
Take the coefficient of \(x\) (which is 6), divide it by 2, and square it:
\[
\left(\frac{6}{2}\right)^2 = 3^2 = 9
\]
Add this value to both sides of the equation:
\[
x^2 + 6x + 9 = -\frac{3}{2} + 9
\]
Simplify the right side:
\[
x^2 + 6x + 9 = \frac{-3 + 18}{2} = \frac{15}{2}
\]
4. **Express the left side as a perfect square:**
\[
(x + 3)^2 = \frac{15}{2}
\]
5. **Solve for \(x\):**
Take the square root of both sides:
\[
x + 3 = \pm \sqrt{\frac{15}{2}}
\]
Simplifying the square root:
\[
x + 3 = \pm \frac{\sqrt{30}}{2}
\]
Finally, solve for \(x\):
\[
x = -3 \pm \frac{\sqrt{30}}{2}
\]
So the solutions for \(x\) are:
\[
x = -3 + \frac{\sqrt{30}}{2} \quad \text{or} \quad x = -3 - \frac{\sqrt{30}}{2}
\]
