How to solve a quadratic equation (with a worked example)

How to · Updated 2026-11-10

Find the discriminant b² − 4ac, then use x = (−b ± √D) / 2a, or split the middle term when the roots are whole numbers or fractions. Worked example and checks.

To solve ax² + bx + c = 0, work out the discriminant D = b² − 4ac first, then put it into x = (−b ± √D) / 2a; when D is a perfect square you can also factorise by splitting the middle term. The quadratic explorer does every one of these steps for the numbers you type and draws the parabola, so you can see the roots as the points where the curve meets the x axis.

What you need

• The equation in standard form, ax² + bx + c = 0, with 0 on one side. For example, 2x² = 7x − 3 becomes 2x² − 7x + 3 = 0.
• The three numbers a, b and c, with their signs. a is the number in front of x², b the number in front of x, and c the number on its own. If a term is missing, its number is 0.
• a must not be 0. With no x² term the equation is linear.

Step by step

1. Open the quadratic explorer and type a, b and c into the boxes beside the sliders. Check the equation printed above the graph for sign slips.
2. Look at the card headed The discriminant decides. It shows D = b² − 4ac with your numbers in it, and a coloured label for what D tells you:
   • D > 0: two different real roots (the curve crosses the x axis twice),
   • D = 0: one repeated root (the lowest or highest point sits on the axis),
   • D < 0: no real roots (the curve never reaches the axis).
3. Under that, the formula line fills in x = (−b ± √D) / 2a and gives both answers.
4. For a factorised answer, scroll to Split the middle term. It looks for two whole numbers that add to b and multiply to a × c and shows the two brackets, or tells you no such pair exists.
5. The Completing the square card lists a third method line by line, with tiles that show where the extra square comes from.

The formula, x = (−b ± √(b² − 4ac)) / 2a, gives two answers: one with plus, one with minus. The part under the root is D, which is why it decides how many real roots there are.

A worked example

Solve 2x² − 7x + 3 = 0.

• Here a = 2, b = −7 and c = 3.
• D = (−7)² − 4 × 2 × 3 = 49 − 24 = 25. That is positive, so there are two real roots, and because 25 = 5² they are rational.
• Formula: x = (7 ± 5) / 4. With plus, x = 12 / 4 = 3. With minus, x = 2 / 4 = 0.5.
• Splitting the middle term: a × c = 6, and the two numbers that add to −7 and multiply to 6 are −6 and −1. So 2x² − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3), which is zero when x = 1/2 or x = 3.
• The graph opens upwards (a is positive), with its lowest point at (1.75, −3.125) and the two green dots at 0.5 and 3.

Now try one that does not factorise: x² − 6x + 7 = 0. Here D = 36 − 28 = 8, which is not a perfect square, so the split card finds no whole-number pair. The formula still works: x = (6 ± √8) / 2 = 3 ± √2, which is about 1.586 and 4.414.

And one with no real answer: x² + 4x + 5 = 0 gives D = 16 − 20 = −4. The curve floats above the axis with its lowest point at (−2, 1). Switch the level to Class 11 and the page gives the complex roots, −2 ± i.

Check your answer

• Put each root back in. For x = 3: 2 × 9 − 21 + 3 = 0. For x = 0.5: 2 × 0.25 − 3.5 + 3 = 0. If you do not get 0, a sign slipped.
• Sum and product. The two roots always add to −b/a and multiply to c/a. Here 3 + 0.5 = 3.5 = 7/2 and 3 × 0.5 = 1.5 = 3/2. Both match, so the pair is right.
• Symmetry. The roots sit the same distance either side of x = −b/2a. 1.75 − 0.5 = 1.25 and 3 − 1.75 = 1.25.

To compare several parabolas on one set of axes, draw them in the graphing calculator; to expand or factorise longer expressions with every step shown, use symbolic algebra.

Questions people ask

Which method should I use in a test? Work out D first. If it is a perfect square, splitting the middle term is quick; if not, go straight to the formula.

Why do I sometimes get the same root twice? When D = 0 the ± part adds and takes away nothing, so both answers are −b/2a. On the graph the curve just touches the axis at its vertex instead of crossing it.

Can a quadratic have only one root when D is positive? No. A positive D always gives two different real roots; you probably forgot the minus version of ±.