For some quadratic x<sup>2</sup> + bx + c = 0,
to complete the square, focus on the x<sup>2</sup> + bx part: we always halve 'b' so we have
(x + (b/2))<sup>2</sup>.
If you expand (x + (b/2))<sup>2</sup>, you have
...(x + (b/2))(x + (b/2)) = x<sup>2</sup> + x(b/2) +x(b/2) + (b<sup>2</sup>/<sup>2</sup>) = x<sup>2</sup> + b<sup>2</sup>/4.
This is something you will get used to with practise.
The next step is to subtract the b<sup>2</sup>/4 term which appears here because we only had x<sup>2</sup> + bx.
The 'c' term finally comes into play with simplification, after having completed the square.
If there was an 'a' infront of x<sup>2</sup> then there's little bit more work to do.
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