[proofplan] We derive the roots of $f(t) = t^2 + at + b$ by completing the square. This rewrites the quadratic as a perfect square minus a constant, reducing the problem to extracting a square root. The technique works because it transforms a degree-2 equation into one that is linear in $(t + a/2)^2$. [/proofplan] [step:Complete the square] [guided] The idea behind completing the square is to absorb the linear term $at$ into a perfect square $(t + a/2)^2 = t^2 + at + a^2/4$. This isolates the unknown $t$ inside a single squared expression, so that the equation $f(t) = 0$ becomes solvable by taking square roots. [/guided] Write $f(t) = t^2 + at + b$. Adding and subtracting $a^2/4$ gives \begin{align*} f(t) &= t^2 + at + \frac{a^2}{4} - \frac{a^2}{4} + b \\ &= \left(t + \frac{a}{2}\right)^2 - \frac{a^2 - 4b}{4}. \end{align*} [/step] [step:Solve for the roots] Setting $f(t) = 0$ and rearranging: \begin{align*} \left(t + \frac{a}{2}\right)^2 = \frac{a^2 - 4b}{4}. \end{align*} Taking square roots in $\mathbb{C}$ yields \begin{align*} t + \frac{a}{2} = \pm\,\frac{\sqrt{a^2 - 4b}}{2}, \end{align*} and therefore \begin{align*} t = \frac{-a \pm \sqrt{a^2 - 4b}}{2}. \end{align*} Since $f$ has degree $2$, it has at most two roots in $\mathbb{C}$, so these are all the roots. Hence $\mathrm{Root}_f(\mathbb{C}) = \left\{ \dfrac{-a + \sqrt{a^2 - 4b}}{2},\; \dfrac{-a - \sqrt{a^2 - 4b}}{2} \right\}$. $\blacksquare$ [/step]