Deriving the quadratic formula

Step 1 (Normalize). Since $a\neq0$, divide every term of $ax^2+bx+c=0$ by $a$: this gives $x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$, an equivalent equation where the leading coefficient is $1$.

Step 2 (Complete the square). The first two terms $x^2+\dfrac{b}{a}x$ are the start of the perfect square $\left(x+\dfrac{b}{2a}\right)^2=x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}$. Adding and subtracting $\dfrac{b^2}{4a^2}$ inside $x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$ lets us rewrite it as $\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=0$.

Step 3 (Isolate the square). Move the constant terms to the right side and combine them over the common denominator $4a^2$: $-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=\dfrac{-b^2+4ac}{4a^2}=-\dfrac{b^2-4ac}{4a^2}=-\dfrac{\Delta}{4a^2}$. This gives $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$.

Step 4 (Take the square root). When $\Delta\ge0$, the right side is a nonnegative real number, so both sides have real square roots: $x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{\Delta}}{2a}$ (the $\pm$ already accounts for both signs, whichever sign $a$ itself has).

Step 5 (Isolate x). Subtract $\dfrac{b}{2a}$ from both sides: $x=-\dfrac{b}{2a}\pm\dfrac{\sqrt{\Delta}}{2a}=\dfrac{-b\pm\sqrt{\Delta}}{2a}$, which is exactly $x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$. Every algebraic step used (dividing by nonzero $a$, adding/subtracting the same quantity, taking square roots of equal nonnegative numbers) is reversible, so this formula is both necessary and sufficient whenever $\Delta\ge0$.