➗ Quadratic Formula

Solve equations of the form \(ax^2 + bx + c = 0\)

Method

Difficulty

Click "New Equation" to begin

Enter the two solutions (order doesn't matter). If there is one repeated solution, enter the same value twice. Fractions like 3/2 are fine.

x₁ =

x₂ =

Score: 0/0

The Quadratic Formula

Any quadratic equation written as \(ax^2 + bx + c = 0\) (where \(a \neq 0\)) can be solved with this formula:

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

The Coefficients

\(a\) = number in front of \(x^2\) \(b\) = number in front of \(x\) \(c\) = the constant term

The Discriminant

\(\Delta = b^2 - 4ac\)

The part under the square root tells you how many real solutions there are.

\(\Delta > 0\)

Two real solutions

The \(\pm\) gives two different answers.

\(\Delta = 0\)

One repeated solution

\(x = \dfrac{-b}{2a}\)

\(\Delta < 0\)

No real solutions

The square root of a negative number is not a real number.

Vertex (turning point)

\(x = \dfrac{-b}{2a}\)

The lowest or highest point of the parabola.

Worked Example

Solve \(x^2 - 5x + 6 = 0\)

1. Identify the coefficients: \(a = 1\), \(b = -5\), \(c = 6\)

2. Discriminant: \(\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1\)

3. Square root: \(\sqrt{1} = 1\)

4. Apply the formula: \(x = \dfrac{5 \pm 1}{2}\)

5. Two solutions: \(x = \dfrac{6}{2} = 3\) and \(x = \dfrac{4}{2} = 2\)

Tips

Watch the signs! A negative \(b\) becomes positive when you compute \(-b\).
Always rearrange the equation so one side equals 0 first.
If \(b^2 - 4ac\) is a perfect square, the answers will be nice whole numbers or fractions.
You can check your answers by substituting them back into the original equation.

📈 Graph Explorer

Drag the sliders to change \(a\), \(b\) and \(c\) and watch the parabola \(y = ax^2 + bx + c\) change shape. Adjust the \(x\)-range sliders to zoom in or out.

x min -10

x max 10

Solving by Factoring

Many quadratics can be solved without the big formula. The idea is to rewrite \(x^2 + bx + c = 0\) as a product of two brackets:

\[ x^2 + bx + c = (x + p)(x + q) = 0 \]

Then, because a product is zero only when one of its factors is zero, the solutions are \(x = -p\) or \(x = -q\).

The Key Trick

\(p + q = b\) \(p \times q = c\)

Find two numbers that add to \(b\) and multiply to \(c\).

Zero-Product Rule

If \(A \times B = 0\) then \(A = 0\) or \(B = 0\)

This is why factoring solves the equation.

The Solutions

\(x + p = 0 \Rightarrow x = -p\) \(x + q = 0 \Rightarrow x = -q\)

Flip the sign of each bracket number.

⚠️ Limitations

Factoring is fast and neat, but it doesn't work for every quadratic:

It only works easily when \(c\) splits into two whole numbers that also add to \(b\). If no such pair exists, you can't factor with simple integers.
The roots must be rational. Equations like \(x^2 - 2 = 0\) (roots \(\pm\sqrt{2}\)) cannot be factored with whole numbers.
If the discriminant \(b^2 - 4ac\) is negative there are no real solutions, so there is nothing to factor.
When the leading coefficient is not 1 (\(a \neq 1\)), factoring takes extra steps. The factoring practice keeps \(a = 1\) so the method always works.

When factoring fails, fall back on the quadratic formula — it always works.

Worked Example

Solve \(x^2 + 5x + 6 = 0\)

1. We need two numbers that multiply to \(6\) and add to \(5\).

2. Those numbers are \(2\) and \(3\) (since \(2 \times 3 = 6\) and \(2 + 3 = 5\)).

3. Factor: \((x + 2)(x + 3) = 0\)

4. Set each factor to zero: \(x + 2 = 0\) or \(x + 3 = 0\)

5. Solutions: \(x = -2\) and \(x = -3\)

Try it yourself on the 🎯 Practice Problems tab — pick the ✏️ Factoring method.