You can solve quadratic equations online for free in seconds: open the Quadratic Calculator, enter the three coefficients, and read both roots from the formula x = (−b ± √(b²−4ac))/2a. For example, x² − 5x + 6 = 0 has a discriminant of 1 and two roots, x = 3 and x = 2. This guide explains every part of that formula, so the calculator result is something you understand and can check, not just copy.
A quadratic equation is any equation that can be arranged into the form ax² + bx + c = 0, where a is not zero. The squared term is what makes it quadratic, and it is also what makes the topic richer than linear equations: a quadratic can have two solutions, one repeated solution, or no real solution at all, and you cannot tell which case you are in until you look at one key quantity, the discriminant. Students meet quadratics in algebra courses, and the same equations reappear wherever a quantity rises and then falls — a thrown ball, a revenue curve, an area problem with one unknown side. This guide works through the formula slowly, compares it honestly with factoring, and uses three anchor equations that between them cover all three possible outcomes, so that any new equation you meet can be placed against a case you have already seen solved in full.
How Do You Solve a Quadratic Equation Online, Step by Step?
- Open the calculator. Go to the Quadratic Calculator with the equation written in front of you.
- Arrange the equation as ax² + bx + c = 0. Move every term to one side so the other side is zero. This step matters because the coefficients only mean what the formula expects when the equation is in this standard form.
- Read off a, b and c, with their signs. In x² − 5x + 6 = 0, a is 1, b is −5 (the minus sign belongs to b), and c is 6. A missing term means a coefficient of zero: x² − 9 = 0 has b = 0.
- Enter the three coefficients. The calculator evaluates the discriminant and both branches of the formula.
- Read the discriminant before the roots. A positive value promises two distinct real roots, zero promises one repeated root, and a negative value means there are no real roots to report.
- Check at least one root by substitution. Put the value back into the original equation and confirm both sides balance. Thirty seconds of checking catches nearly every sign error.
The Quadratic Formula and the Discriminant
The formula x = (−b ± √(b²−4ac))/2a solves every quadratic equation, without exception, provided a is not zero. It is usually presented as one object, but it is easier to trust when it is taken apart. The −b at the front recentres the equation around its axis of symmetry. The denominator 2a scales the answer for how steeply the curve opens. The ± symbol is the reason quadratics can have two answers: one calculation adds the square-root term and the other subtracts it, and both results are genuine solutions unless the square-root term happens to be zero. Nothing in the formula is decoration, and the part that repays the closest attention sits under the root sign.
That part, b² − 4ac, is called the discriminant, and its sign alone classifies the equation completely. If b² − 4ac is positive, its square root is a positive real number, the two branches separate, and the equation has two distinct real roots. If it is exactly zero, the square root vanishes, both branches collapse onto −b/2a, and there is one repeated root. If it is negative, the square root has no real value, and the equation has no real roots at all — its graph, a parabola, floats entirely above or below the horizontal axis without ever touching it. This is why experienced solvers compute the discriminant first: it is a short calculation that tells you what shape of answer is coming, and it turns a surprising result — a root that will not check, a graph with no crossing — into an expected one.
| Method or case | Condition | Worked check |
|---|---|---|
| Discriminant positive | b² − 4ac > 0 | x² − 5x + 6 = 0: D = 1, roots 3 and 2 |
| Discriminant zero | b² − 4ac = 0 | x² − 4x + 4 = 0: D = 0, repeated root 2 |
| Discriminant negative | b² − 4ac < 0 | x² + x + 1 = 0: D = −3, no real roots |
| Factoring | Two numbers multiply to c and add to b (when a = 1) | −2 and −3: product 6, sum −5 |
| Formula | Works for every a, b and c with a ≠ 0 | x = (−b ± √D) / 2a |
Factoring Versus the Formula: An Honest Comparison
Factoring rewrites the quadratic as a product, such as (x − 2)(x − 3) = 0, and then uses the fact that a product is zero only when one of its factors is zero, giving x = 2 and x = 3 directly. When the factors are whole numbers and easy to spot, factoring is quicker, cleaner, and shows the structure of the equation in a way the formula hides. Its weakness is that it only works conveniently when the roots are tidy. Many quadratics have roots involving square roots or fractions, and some have no real roots at all; staring at such an equation, hunting for factors that do not exist in a usable form, wastes time. The formula has the opposite profile: it is longer, it invites sign slips, and it always works. A sound habit is to glance for factors for a few seconds, factor when they are obvious, and move to the formula without hesitation when they are not — using the discriminant in both cases to confirm how many roots the factoring or the formula ought to find.
Worked Examples: All Three Outcomes
Example 1 — two distinct roots. Solve x² − 5x + 6 = 0. Here a = 1, b = −5 and c = 6. The discriminant is b² − 4ac = 25 − 24 = 1, which is positive, so expect two distinct real roots. The formula gives x = (5 ± √1) / 2 = (5 ± 1) / 2. Adding gives 6 / 2 = x = 3; subtracting gives 4 / 2 = x = 2. Factoring agrees: (x − 2)(x − 3) expands back to the original equation. Checking x = 3: 9 − 15 + 6 = 0, correct.
Example 2 — one repeated root. Solve x² − 4x + 4 = 0. Here a = 1, b = −4 and c = 4. The discriminant is 16 − 16 = 0, so expect a single repeated root. The formula gives x = (4 ± 0) / 2 = x = 2, the same value from both branches. Factoring shows why: the expression is (x − 2)(x − 2), a perfect square, so the factor that produces the root appears twice. Checking: 4 − 8 + 4 = 0, correct.
Example 3 — no real roots. Solve x² + x + 1 = 0. Here a = 1, b = 1 and c = 1. The discriminant is 1 − 4 = −3, which is negative, so there are no real roots. This is a complete and correct answer, not a failed calculation: the parabola y = x² + x + 1 stays above the axis everywhere, since its lowest value is still positive. Recognising this case quickly matters, because without the discriminant check it is easy to burn time hunting for factors or roots that do not exist on the real number line.
Discriminant values in this guide and what each one means — hover or tap each bar for details.
The chart puts the three anchor discriminants side by side, and their relative sizes are less important than their signs. The bar for D = 1 is positive and produces two roots a whole unit apart after the square root is taken. The bar for D = 0 is barely there at all, matching a square-root term that contributes nothing and leaves a single root. The bar for D = −3 sits on the negative side of the story: it cannot be square-rooted within the real numbers, so the equation stops there. Whenever a new quadratic lands in front of you, placing its discriminant on this small picture — clearly positive, exactly zero, or negative — is the fastest reliable summary of what its solutions will look like.
Where Quadratic Equations Show Up
| Situation | What the roots mean | Watch for |
|---|---|---|
| Projectile and falling-object problems | The times when the object is at a given height | Two positive times can both be meaningful; a negative time usually is not |
| Area problems with one unknown side | Possible lengths that give the stated area | Lengths must be positive, so a negative root is discarded on meaning, not on arithmetic |
| Revenue and profit models | Prices or quantities where profit is zero or at a stated level | Both roots can be real and relevant, marking the edges of a profitable range |
| Algebra coursework | The solutions themselves | Show the discriminant step; it justifies the number of roots you report |
| Geometry with squared lengths | A length satisfying a squared relationship | Check the root against the diagram before accepting it |
Five Mistakes That Spoil Quadratic Solutions
- Losing the sign of b. In x² − 5x + 6 = 0, b is −5, not 5. The formula starts with −b, so a sign dropped at the reading-off stage flips the working while looking perfectly tidy. Write a, b and c down, signs included, before touching the formula.
- Squaring a negative incorrectly. b² means b multiplied by itself, so (−5)² is 25, never −25. Most discriminant errors are born here, one line before the formula itself is even used.
- Dividing only part of the top line by 2a. The denominator 2a applies to the whole of −b ± √D, not just to the square-root term. Splitting the fraction incorrectly changes both roots.
- Stopping at one root. The ± in the formula is an instruction to calculate twice. Reporting only the added branch throws away a genuine solution whenever the discriminant is positive.
- Forcing an answer when the discriminant is negative. A negative discriminant is the answer: no real roots. Inventing a value by ignoring the minus sign under the root produces numbers that will never check by substitution.
Conclusion: Let the Discriminant Speak First
Every quadratic yields to the same short routine: write it as ax² + bx + c = 0, read off the coefficients with their signs, compute b² − 4ac, and let its sign tell you whether to expect two roots, one repeated root, or none. Factor when the factors are obvious, use the formula for everything else, and check a root by substitution before trusting it. The Quadratic Calculator carries out the arithmetic instantly once the coefficients are right. Companion tools cover the arithmetic quadratics lean on: the fraction calculator for fractional coefficients and roots, and the average calculator for the mean of the two roots, which is always −b/2a. For background reading, our guide to calculating fractions covers the fraction skills the formula demands, and our guide to calculating standard deviation shows square roots and squared terms at work in a different setting.
School, tutoring and course websites lose enquiries when a worked example contains a wrong root that a student spots before the teacher does. If your site publishes maths help, worksheets or course pages that families rely on, book a free consultation with Asad Webs and mention the pages involved; clear, checked worked examples are a small fix that protects the trust every enrolment depends on.
Frequently Asked Questions
What is the quadratic formula?
For an equation written as ax² + bx + c = 0, the solutions are x = (−b ± √(b² − 4ac)) / 2a. The letter a is the coefficient of x², b is the coefficient of x, and c is the constant term. The ± symbol means the formula produces two calculations: one adding the square root and one subtracting it.
What does the discriminant tell you?
The discriminant is the part under the square root, b² − 4ac. If it is positive, the equation has two distinct real roots. If it is zero, there is one repeated real root. If it is negative, there are no real roots, because no real number squares to a negative value. Checking it first tells you what kind of answer to expect before you finish the calculation.
When should I factor instead of using the formula?
Factor first when the factors are easy to spot — small whole-number coefficients where two numbers multiply to c and add to b. Factoring x² − 5x + 6 into (x − 2)(x − 3) is faster than the formula. Use the formula when factoring is not obvious, when coefficients are large or fractional, or when you suspect there may be no real roots at all.
Can a quadratic equation have only one solution?
Yes. When the discriminant is exactly zero, the plus and minus branches of the formula give the same value, so the two roots coincide. The equation x² − 4x + 4 = 0 has the single repeated root x = 2, and its graph touches the horizontal axis at one point instead of crossing it twice.
What if a is zero?
Then the equation is not quadratic at all. If a is zero, the x² term disappears and you are left with a linear equation, bx + c = 0, which is solved by moving c across and dividing by b. The quadratic formula also breaks in that case, because it divides by 2a, and division by zero is undefined.
Why can a quadratic have no real roots?
Because solving it can require the square root of a negative number. For x² + x + 1 = 0 the discriminant is −3, and no real number squared gives −3. The equation still has solutions in the complex number system, but on the real number line, and on an ordinary graph, its curve never reaches the axis.
How do I check a root is correct?
Substitute it back into the original equation. For x = 3 in x² − 5x + 6 = 0: 9 − 15 + 6 = 0, which is true. For x = 2: 4 − 10 + 6 = 0, also true. A value that does not make the original equation balance is not a root, however neatly it came out of the formula.
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