A Complete Proof

Deriving the Quadratic Formula

By completing the square on the general quadratic equation

Why This Matters

Every quadratic equation of the form $y = ax^2 + bx + c$ can be solved for its roots — the $x$-values where the parabola crosses the $x$-axis. To find those roots, we set $y = 0$ and solve:

We start with
$$ax^2 + bx + c = 0 \qquad \text{where } a, b, c \in \mathbb{R} \text{ and } a \neq 0$$

The quadratic formula gives us the roots directly — but it isn't pulled from thin air. It comes from a technique called completing the square, applied to the general equation above. Here is the full derivation, one careful step at a time.

The Derivation

Step 1
Move the constant to the right side

Subtract $c$ from both sides to isolate the terms containing $x$:

$$ax^2 + bx = -c$$
↓
Step 2
Divide everything by $a$

We need the coefficient of $x^2$ to be $1$ before we can complete the square:

$$x^2 + \frac{b}{a}\,x = -\frac{c}{a}$$
↓
Step 3
Identify the coefficient of $x$

The coefficient of the linear term $x$ is $\dfrac{b}{a}$. This number is the key to completing the square.

↓
Step 4
Take half of it, then square it

Divide the coefficient by $2$ and square the result:

$$\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$$
Why this works

We are exploiting the algebraic identity $(x + d)^2 = x^2 + 2dx + d^2$. If we set $2d = \tfrac{b}{a}$, then $d = \tfrac{b}{2a}$, and the "missing piece" to complete the perfect square is $d^2 = \tfrac{b^2}{4a^2}$.

↓
Step 5
Add $\dfrac{b^2}{4a^2}$ to both sides

Whatever we add to one side, we must add to the other to keep the equation balanced:

$$x^2 + \frac{b}{a}\,x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$$
↓
Step 6
Simplify the right side

Combine the two fractions on the right using a common denominator of $4a^2$:

$$-\frac{c}{a} + \frac{b^2}{4a^2} = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}$$

So the equation becomes:

$$x^2 + \frac{b}{a}\,x + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}$$
↓
Step 7
Write the left side as a perfect square

The left side is now a perfect square trinomial — it factors as a binomial squared:

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$
Verify it

Expand $\left(x + \tfrac{b}{2a}\right)^2 = x^2 + 2 \cdot x \cdot \tfrac{b}{2a} + \left(\tfrac{b}{2a}\right)^2 = x^2 + \tfrac{b}{a}\,x + \tfrac{b^2}{4a^2}$. ✓

↓
Step 8
Take the square root of both sides

The square root undoes the square on the left. On the right, we must include $\pm$ because both a positive and negative value could have been squared to produce the right side:

$$x + \frac{b}{2a} = \pm\,\frac{\sqrt{b^2 - 4ac}}{2a}$$
Why ±?

Both $(+3)^2 = 9$ and $(-3)^2 = 9$. Whenever we take a square root in an equation, we account for both possibilities with $\pm$.

↓
Step 9
Isolate $x$

Subtract $\dfrac{b}{2a}$ from both sides:

$$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
↓
Step 10
Combine into a single fraction

Both terms share the denominator $2a$, so we write them as one fraction:

The Quadratic Formula
$$\boxed{\;x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\;}$$

The Discriminant

The expression under the square root, $\,\Delta = b^2 - 4ac\,$, is called the discriminant. It tells you what kind of roots the equation has before you even solve it:

DiscriminantMeaningRoots
$b^2 - 4ac > 0$Positive under the radicalTwo distinct real roots
$b^2 - 4ac = 0$Zero under the radicalOne repeated real root
$b^2 - 4ac < 0$Negative under the radicalTwo complex conjugate roots

Quick Example

Solve $\;2x^2 - 4x - 3 = 0$.

Here $a = 2$, $b = -4$, $c = -3$. Plug into the formula:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-3)}}{2(2)} = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm \sqrt{40}}{4} = \frac{4 \pm 2\sqrt{10}}{4} = \frac{2 \pm \sqrt{10}}{2}$$

So the two roots are $\;x = \dfrac{2 + \sqrt{10}}{2} \approx 2.58\;$ and $\;x = \dfrac{2 - \sqrt{10}}{2} \approx -0.58$.