Let's solve each equation by completing the square.

Problem (a): $x^2 + 6x + 3 = 0$

1. Move the constant to the other side:
   $x^2 + 6x = -3$

2. Take half of the coefficient of $x$ (which is 6), square it, and add it to both sides.
   $\left(\frac{6}{2}\right)^2 = 9$ $x^2 + 6x + 9 = -3 + 9$

3. This gives:
   $(x + 3)^2 = 6$

4. Take the square root of both sides:
   $x + 3 = \pm \sqrt{6}$

5. Solve for $x$:
   $x = -3 \pm \sqrt{6}$

Problem (b): $s^2 + 2s - 6 = 0$

1. Move the constant to the other side:
   $s^2 + 2s = 6$

2. Take half of the coefficient of $s$ (which is 2), square it, and add it to both sides.
   $\left(\frac{2}{2}\right)^2 = 1$ $s^2 + 2s + 1 = 6 + 1$

3. This gives:
   $(s + 1)^2 = 7$

4. Take the square root of both sides:
   $s + 1 = \pm \sqrt{7}$

5. Solve for $s$:
   $s = -1 \pm \sqrt{7}$

Problem (c): $x^2 + 4x - 2 = 0$

1. Move the constant to the other side:
   $x^2 + 4x = 2$

2. Take half of the coefficient of (x