How to Use the Quadratic Formula Step by Step (Worked Examples)

Quick answer: To use the quadratic formula, rearrange your equation into ax^2 + bx + c = 0, read off a, b and c with their signs, then substitute into x = (-b ± sqrt(b^2 - 4ac)) / (2a). Work out the discriminant b^2 - 4ac first, then do the plus case and the minus case separately.

I lost four marks on a mock last term because I wrote b = 7 when the equation said -7x. The method was perfect, the substitution line was perfect, and everything after it was wrong. So this article is less about the formula itself, which takes thirty seconds to learn, and more about the mechanical bits where marks actually leak out. I do IGCSE and A-Level, and I have made every one of these mistakes at least twice.

Getting a, b and c right is the whole game

The formula only works if the equation is in the form ax^2 + bx + c = 0, with zero on the right-hand side, so everything must be moved across before you read off anything.

Then read the coefficients including their signs. In 3x^2 - 7x - 5 = 0, that gives a = 3, b = -7, c = -5. The minus signs belong to the numbers. Write them in a little column before you touch the formula:

• a = 3
• b = -7
• c = -5

That takes five seconds and it is the highest-value habit in this topic. If a term is missing, its coefficient is 0, so in 2x^2 - 18 = 0, b = 0. If x^2 has no number in front, a = 1, not 0.

A full worked example with every line shown

Solve 3x^2 - 7x - 5 = 0, giving answers to 2 decimal places.

1. Coefficients: a = 3, b = -7, c = -5.
2. Discriminant: b^2 - 4ac = (-7)^2 - 4 × 3 × (-5) = 49 - (-60) = 49 + 60 = 109.
3. Square root: sqrt(109) = 10.4403.
4. Numerator: -b = 7, so the top is 7 ± 10.4403.
5. Denominator: 2a = 6.
6. Plus case: x = (7 + 10.4403) / 6 = 17.4403 / 6 = 2.9067.
7. Minus case: x = (7 - 10.4403) / 6 = -3.4403 / 6 = -0.5734.

So x = 2.91 or x = -0.57 to 2 decimal places.

Notice step 2. Because c was negative, -4ac turned into +60, so subtracting a negative added. That one line is where about half of all lost marks in this topic live.

Quick check: put 2.91 back into the original. 3(2.91)^2 = 25.40, then -7(2.91) = -20.37, then -5. Total 0.03, which is just rounding. That check takes fifteen seconds and I now do it on every quadratic in an exam.

When the equation is not in standard form yet

Most exam questions do not hand you ax^2 + bx + c = 0. They hand you something like 2x^2 + 5 = 9x - 1.

Move everything to the left: 2x^2 - 9x + 6 = 0. So a = 2, b = -9, c = 6.

Discriminant: 81 - 4 × 2 × 6 = 81 - 48 = 33. Positive but not a square number, so the answers will be surds or decimals.

x = (9 ± sqrt(33)) / 4

sqrt(33) = 5.7446, so x = 14.7446 / 4 = 3.69 or x = 3.2554 / 4 = 0.81 to 2 decimal places.

With a worded set-up, an area, or a fraction equation, expand and clear denominators first, then collect on one side. Nothing goes into the formula until the right-hand side is a bare zero. If rearranging is shaky, our maths solver shows those steps rather than just the answer.

What the discriminant tells you before you solve

The discriminant is b^2 - 4ac, the bit under the square root. On its own it answers "how many real solutions?" without you solving anything.

• Positive: two different real roots. The parabola crosses the x-axis twice.
• Zero: one repeated root. The parabola touches the x-axis at exactly one point.
• Negative: no real roots. The parabola never reaches the x-axis.

Worked example of the zero case: x^2 - 6x + 9 = 0 gives 36 - 36 = 0, so x = 6/2 = 3, a repeated root.

Worked example of the negative case: x^2 + 2x + 5 = 0 gives 4 - 20 = -16. Negative, so no real solutions, and your calculator will throw a math error if you try.

Mark-scheme wording matters here. If a question says "show that this equation has no real roots", a full-mark answer computes b^2 - 4ac, states the value, says it is negative, then concludes "therefore no real roots". Writing only "-16" earns the method mark and loses the conclusion mark. CBSE Class 10 asks for the "nature of the roots" in almost exactly these words, so learn the phrase.

Exact surd form versus rounded decimals

Read the instruction at the end of the question before you round anything.

If it says "to 2 decimal places" or "to 3 significant figures", that is a strong hint the equation does not factorise, so keep full precision until the final line. Rounding sqrt(109) to 10.4 early can shift your answer enough to lose the accuracy mark.

If it says "give your answers in exact form" or "leave your answer as a surd", do not touch the decimal button. Simplify instead. For example x = (4 ± sqrt(48)) / 2 becomes x = (4 ± 4sqrt(3)) / 2, and every term divides by 2, so x = 2 ± 2sqrt(3). Simplify the surd first, then cancel, and only cancel if every term on top shares the factor.

One more thing worth knowing: the substitution line earns its own method mark. Even if your arithmetic collapses afterwards, writing x = (7 ± sqrt((-7)^2 - 4 × 3 × (-5))) / (2 × 3) in full banks a mark. Never do the substitution in your head.

Formula or factorising? How I decide in five seconds

Factorising is faster when it works, so I give it one quick look. If a = 1 and I can spot two numbers that multiply to c and add to b almost immediately, I factorise. If not, I stop hunting and use the formula.

Three tells that it will not factorise nicely:

• The question specifies decimal places or significant figures.
• The discriminant is not a square number.
• a is something awkward like 6 or 7 with a large c.

The formula never fails; factorising fails often. In a timed paper, thirty wasted seconds hunting for factors of 3x^2 - 7x - 5 costs more than just running the formula. This turns up across Cambridge IGCSE, GCSE higher tier with AQA, Edexcel and OCR, WASSCE core maths, and AP work where quadratics sit inside bigger problems. Drill twenty of these with a generated quiz, or work through the wider topic on the maths hub.

Test yourself

1. Solve 2x^2 + 7x - 3 = 0, giving your answers to 2 decimal places.
2. Without solving, state how many real roots x^2 - 10x + 25 = 0 has, and justify your answer.
3. Rearrange and solve 5x^2 = 3x + 4, leaving your answer in exact surd form.

Answers: 1. x = 0.39 or x = -3.89. 2. Discriminant = 0, so one repeated root, x = 5. 3. x = (3 ± sqrt(89)) / 10.

For fully worked exam-style quadratic word problems, see quadratic word problems solved step by step.

FAQ

Do I need to memorise the quadratic formula?

Some papers print it in the formulae list at the front and some do not, and boards change their sheets, so check the current one for your board. Memorise it regardless. Flipping to the front page mid-question breaks your concentration, and you will use this formula for years.

Can I use the quadratic formula even when the equation factorises?

Yes. It works on every quadratic, factorisable or not, and you will not be penalised for choosing it. Factorising is only ever a speed choice, never a correctness one.

Why does my calculator say math error?

Almost always a negative discriminant, which means there are no real solutions. Check your signs first, because a dropped minus is more likely than a genuinely negative discriminant. If the signs are right, "no real roots" is the answer.

What if a is negative, like -2x^2 + 5x + 3 = 0?

The formula still works exactly as written. If negative signs make you nervous, multiply the whole equation by -1 first to get 2x^2 - 5x - 3 = 0. The solutions are identical and the arithmetic is friendlier. If a step still will not click, ask for it broken down on the explain page.

In short: rearrange to equal zero, list a, b and c with their signs, work out the discriminant on its own line, then split into the plus and minus cases. Write the substitution out in full, keep full precision until the final rounding, and check whether the question wants surds or decimals. Do that consistently and the quadratic formula becomes free marks rather than a place where they quietly disappear.