Completing the square works by noticing that

$$(x+a)^2 = x^2 + 2ax + a^2$$

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

a)

Rewrite $x^2+\var{sml}x$ as $\simplify[basic]{ (x+{sml/2})^2 - {sml/2}^2}$.

$$\begin{align} \simplify[basic]{x^2+{sml}x+{big}} &= \simplify[basic]{(x+{sml/2})^2-{(sml/2)}^2+{big}} \\ &= \simplify[basic]{(x+{sml/2})^2+{-(sml/2)^2+big}} \text{.} \end{align}$$

b)

We showed above that

$$\simplify[basic]{x^2+{sml}x+{big}} = 0$$

is equivalent to

$$\simplify[basic]{(x+{bits[0]})^2-{bits[1]^2}} = 0 \text{.}$$

We can then rearrange this equation to solve for $x$.

$$\begin{align} \simplify{(x+{bits[0]})^2-{(bits[1])^2} } &= 0 \\ (x+\var{bits[0]})^2 &= \var{bits[1]^2} \\ x+\var{bits[0]} &= \pm \var{bits[1]} \\ x &= -\var{bits[0]} \pm \var{bits[1]} \\[2em] x_1 &= \var{-bits[0]-bits[1]} \text{,}\\ x_2 &= \var{-bits[0]+bits[1]} \text{.} \end{align}$$