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Algebra

Quadratic formula tutorial

A quadratic equation has the form $ax^2 + bx + c = 0$. The quadratic formula solves any such equation — no matter how hard it is to factor — by plugging the coefficients directly into one formula.

The quadratic formula

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

where $a$, $b$, and $c$ are the coefficients from $ax^2 + bx + c = 0$ (and $a \neq 0$).

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Identifying $a$, $b$, and $c$

Before using the formula, align the equation with the standard form $ax^2 + bx + c = 0$ and read off the three coefficients.

Example: $x^2 - 5x + 6 = 0$

$a = 1$, $b = -5$, $c = 6$

Example: $2x^2 + 3x - 2 = 0$

$a = 2$, $b = 3$, $c = -2$

Example: $x^2 - 9 = 0$ (no $x$ term)

$a = 1$, $b = 0$, $c = -9$

The discriminant: $b^2 - 4ac$

The expression under the square root is called the discriminant. Its value tells you how many real solutions the equation has — before you do the full calculation.

Positive discriminant

$b^2 - 4ac > 0$: two distinct real roots (the $\pm$ gives two different answers).

Zero discriminant

$b^2 - 4ac = 0$: exactly one real root (a "repeated" root).

Negative discriminant

$b^2 - 4ac < 0$: no real roots (solutions involve imaginary numbers — beyond standard Algebra 1).

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Worked example 1

Solve $x^2 - 5x + 6 = 0$.

Identify: $a = 1$, $b = -5$, $c = 6$

Discriminant: $(-5)^2 - 4(1)(6) = 25 - 24 = 1$

Apply formula: $x = \dfrac{-(-5) \pm \sqrt{1}}{2(1)} = \dfrac{5 \pm 1}{2}$

Two roots: $x = \dfrac{5 + 1}{2} = 3$ and $x = \dfrac{5 - 1}{2} = 2$

Solutions: $x = 2$ and $x = 3$

Check: $(3)(2)=6$ ✓ and $3+2=5$ ✓

Worked example 2

Solve $x^2 + x - 12 = 0$.

Identify: $a = 1$, $b = 1$, $c = -12$

Discriminant: $1^2 - 4(1)(-12) = 1 + 48 = 49$

Apply formula: $x = \dfrac{-1 \pm \sqrt{49}}{2} = \dfrac{-1 \pm 7}{2}$

Two roots: $x = \dfrac{-1 + 7}{2} = 3$ and $x = \dfrac{-1 - 7}{2} = -4$

Solutions: $x = -4$ and $x = 3$

Step-by-step checklist

  1. Rearrange the equation to standard form $ax^2 + bx + c = 0$ if needed.
  2. Identify $a$, $b$, and $c$ — note the signs carefully.
  3. Compute the discriminant: $D = b^2 - 4ac$.
  4. If $D < 0$: no real solutions. If $D \geq 0$: continue.
  5. Take $\sqrt{D}$.
  6. Compute both roots: $x = \dfrac{-b + \sqrt{D}}{2a}$ and $x = \dfrac{-b - \sqrt{D}}{2a}$.
  7. Verify by substituting each root back into the original equation.

Try a sample problem

Each practice problem has two roots. Enter the smaller root in the first box. Click Check answer to see the worked solution.