Completing the square works on every quadratic. Do it once with letters instead of numbers and you get a formula that solves them all. It uses completing the square and simplifying radicals.
Why it works
Start with $ax^2 + bx + c = 0$, $a \ne 0$. Divide by $a$: $x^2 + \frac{b}{a}x = -\frac{c}{a}$. Half of $\frac{b}{a}$ is $\frac{b}{2a}$; add its square to both sides: $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$$ Take square roots: $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$. Subtract $\frac{b}{2a}$, and the quadratic formula appears. Every step is a completing-the-square step; nothing is new except the letters.
Key idea
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ solves $ax^2 + bx + c = 0$. The whole numerator, $-b \pm \sqrt{b^2 - 4ac}$, is divided by $2a$.
Worked examples
Worked example: Rational roots
Solve $2x^2 + 3x - 5 = 0$.
- $a = 2$, $b = 3$, $c = -5$. Discriminant: $3^2 - 4(2)(-5) = 9 + 40 = 49$.
- $x = \frac{-3 \pm 7}{4}$.
- $x = \frac{4}{4} = 1$ or $x = \frac{-10}{4} = -\frac{5}{2}$.
Answer: $x = 1$ or $x = -\frac{5}{2}$
Worked example: Irrational roots
Solve $x^2 - 4x - 3 = 0$.
- $a = 1$, $b = -4$, $c = -3$. $b^2 - 4ac = 16 + 12 = 28$.
- $x = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2}$.
- Divide both terms of the numerator by $2$: $x = 2 \pm \sqrt{7}$.
Answer: $x = 2 \pm \sqrt{7}$
Worked example: Standard form first
Solve $3x^2 = 5x + 2$.
- The formula needs $ax^2 + bx + c = 0$. Subtract $5x$ and $2$: $3x^2 - 5x - 2 = 0$, so $a = 3$, $b = -5$, $c = -2$.
- $b^2 - 4ac = 25 + 24 = 49$.
- $x = \frac{5 \pm 7}{6}$: $x = 2$ or $x = -\frac{1}{3}$.
Answer: $x = 2$ or $x = -\frac{1}{3}$
Common mistake
Two common slips: dividing only the square root by $2a$, and computing $-b^2$ when $b$ is negative. The fraction bar covers $-b$ too. And $b^2$ means $(-4)^2 = 16$, never $-16$: write $b$ in parentheses when you substitute. Check one root in the original equation.