Quadratic Formula: How to Solve Any Quadratic Equation Step by Step

The Quadratic Formula

Every quadratic equation of the form

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

has solutions given by the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This single formula works for every quadratic equation — whether it has two real roots, one repeated root, or two complex roots.

Understanding the Discriminant

The expression under the square root is called the discriminant:

$$\Delta = b^2 - 4ac$$

The discriminant tells you the nature of the roots before you even solve:

• $\Delta > 0$ — Two distinct real roots
• $\Delta = 0$ — One repeated real root (a perfect square)
• $\Delta < 0$ — Two complex conjugate roots

Worked Examples

Example 1: Two Real Roots

Solve $2x^2 - 7x + 3 = 0$.

Here $a = 2$, $b = -7$, $c = 3$. First, compute the discriminant:

$$\Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25$$

Since $\Delta = 25 > 0$, we get two distinct real roots:

$$x = \frac{-(-7) \pm \sqrt{25}}{2 \cdot 2} = \frac{7 \pm 5}{4}$$ $$x_1 = \frac{7 + 5}{4} = 3, \qquad x_2 = \frac{7 - 5}{4} = \frac{1}{2}$$

Example 2: Repeated Root

Solve $x^2 - 6x + 9 = 0$.

$$\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$$

One repeated root:

$$x = \frac{6}{2} = 3$$

This makes sense because $x^2 - 6x + 9 = (x - 3)^2$.

Example 3: Complex Roots

Solve $x^2 + 2x + 5 = 0$.

$$\Delta = 4 - 20 = -16$$

Since $\Delta < 0$:

$$x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

The roots are $x = -1 + 2i$ and $x = -1 - 2i$.

Deriving the Formula: Completing the Square

Where does the quadratic formula come from? Start with the general equation and complete the square:

$$ax^2 + bx + c = 0$$

Divide both sides by $a$:

$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

Move the constant term:

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

Add $\left(\frac{b}{2a}\right)^2$ to both sides to complete the square:

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

The left side is now a perfect square:

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

Take the square root of both sides:

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

Solve for $x$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Vieta’s Formulas

For a quadratic $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$:

$$x_1 + x_2 = -\frac{b}{a}, \qquad x_1 \cdot x_2 = \frac{c}{a}$$

These relationships are useful for checking your answers. For Example 1 above: $3 + \frac{1}{2} = \frac{7}{2} = -\frac{-7}{2}$ and $3 \cdot \frac{1}{2} = \frac{3}{2} = \frac{3}{2}$. Both check out.

Vertex Form and the Parabola

A quadratic function $f(x) = ax^2 + bx + c$ graphs as a parabola. The vertex is at:

$$\left(-\frac{b}{2a},\; f\!\left(-\frac{b}{2a}\right)\right)$$

Notice that $-\frac{b}{2a}$ is exactly the midpoint of the two roots — the axis of symmetry of the parabola passes right between them.

When to Use the Quadratic Formula

The quadratic formula always works, but it’s not always the fastest method:

• Factoring is quicker when the roots are “nice” integers or simple fractions.
• Completing the square is useful when you need the vertex form.
• The quadratic formula is the reliable fallback when factoring isn’t obvious.

Quick Reference

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