The quadratic formula $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ for solving the quadratic equation

(1)

Proof. Multiplying (1) by $4a$ and adding $b^2$ to both sides gives an equivalent equation $$4a^2x^2+4abx+4ac+b^2 = b^2$$ or $$(2ax)^2+2\cdot2ax\cdot{b}+b^2 = b^2-4ac$$ or furthermore $$(2ax+b)^2 = b^2-4ac.$$ Taking square root algebraically yields $$2ax+b = \pm\sqrt{b^2-4ac},$$ which implies the quadratic formula.

Note. A similar quadratic formula is meaningful besides $\mathbb{C}$ also in other fields with characteristic $\neq 2$ if one can find the needed ``square root'' (this may require a field extension).