Completing the square Calculator
Completing the square turns a quadratic like 2x² − 12x + 7 into 2(x − 3)² − 11, and that one rewrite hands you the vertex, the axis of symmetry, and a route to the roots that needs nothing but a square root. Enter the three coefficients and this calculator shows the completed square form, every working step, a graph of your parabola, and the geometric area diagram that explains why the method has its name.
The equation is read as ax² + bx + c = 0. Decimals and negative coefficients are welcome; only a = 0 is refused, with the reason shown.
Quick Answer
Key takeaways
- The whole method is one move: add and subtract (b/2a)² so the x terms become a perfect square, then tidy up the constants.
- Vertex form a(x - h)² + k puts the vertex at (h, k). Watch the sign: (x - 3)² means h = +3, not -3.
- The discriminant b² - 4ac predicts the roots before you solve: positive means two real roots, zero means one repeated root, negative means a complex pair.
- Completing the square is not a rival to the quadratic formula; it is the derivation of it, and the only route that also yields the vertex.
- Quick Answer
- How to use the completing the square calculator
- The method in four moves
- Worked example: 2x² − 12x + 7, exactly as the tool prints it
- Reading the graph and the area diagram
- Completing the square or the quadratic formula?
- What competing calculators miss
- Frequently Asked Questions
- Add This Calculator to Your Website
- References
How to use the completing the square calculator
Read your quadratic in the form ax² + bx + c and type the three numbers into a, b, and c, keeping their signs. If your equation has the terms in a different order, or something on the right-hand side, rearrange it to equal zero first; tidying an expression into standard form is the same skill as using a combine like terms calculator on the pieces. Press Complete the Square and the results appear in teaching order: the vertex form first, then the vertex, axis, discriminant, and roots, then the written steps, then the two pictures.
Change one coefficient at a time if you are learning the method. Doubling b moves the vertex sideways and changes the number that gets squared; changing c lifts the whole parabola without moving the axis; making a negative flips the curve so the vertex becomes a maximum. The graph redraws from the same computed values every time, so what you see is always the quadratic you typed, never a stock illustration.
The method in four moves
Every completion follows the same four moves, and the calculator's step list is these moves with your numbers in them.
Move one: if a is not 1, factor it out of the x terms only, leaving c outside the brackets. Move two: inside the brackets, halve the x coefficient and square the result. Move three: add and subtract that square inside the brackets, which changes nothing overall but lets the first three terms collapse into (x - h)². Move four: multiply the subtracted piece back out through a and combine it with c to get k. Lesson sequences such as the Khan Academy completing the square lessons teach the identical sequence, because there is only one: the algebra forces it.
Worked example: 2x² − 12x + 7, exactly as the tool prints it
The calculator is pre-filled with this quadratic. Its step list, verbatim in substance, runs:
| Step | Working |
|---|---|
| Start | 2x² − 12x + 7 = 0 |
| Factor 2 out of the x terms | 2(x² − 6x) + 7 = 0 |
| Halve -6 and square it | Half is -3, and squaring it gives 9 |
| Add and subtract 9 inside | 2(x² − 6x + 9 − 9) + 7 = 0 |
| Collapse the perfect square | 2(x − 3)² − 11 = 0 |
| Solve | (x − 3)² = 5.5, so x = 5.3452 or x = 0.6548 |
Check the vertex against the formulas: h = −(-12) / (2 x 2) = 3, and k = 7 − 144 / 8 = 7 − 18 = −11. Vertex (3, −11), axis x = 3, exactly as reported. The discriminant, 144 − 56 = 88, is positive, which is why two real roots exist and why the graph crosses the axis twice, at 0.6548 and 5.3452, symmetrically placed around x = 3. Expanding 2(x − 3)² − 11 back out gives 2x² − 12x + 18 − 11 = 2x² − 12x + 7, so the rewrite is exact, not an approximation.
Reading the graph and the area diagram
The graph is drawn from your numbers: the vertex is dotted in red, the axis of symmetry is the dashed gold line, and real roots are marked where the curve crosses the axis. When the discriminant is negative there are no crossing points to mark, and the whole curve sits on one side of the axis, which is the picture that makes "no real roots" stop being an abstraction. The same midpoint idea, that the vertex sits halfway between the roots, is the fastest way to sanity-check a hand solution.
The area diagram shows the geometry the method is named after. x² is a square, the bx term is two rectangles stuck to its sides, and the dashed corner is the missing piece: literally the square that has to be completed. For the worked example the added corner is (−3)² = 9 inside the brackets. Once students see the corner, the "half it and square it" rule stops being a spell and becomes a measurement. Diagrams like this are also why the method survives in courses long after formula-only solving would do: the picture generalises to circle equations and to the integrals where completing the square reappears in calculus, a setting explored further by the antidifferentiation calculator.
Completing the square or the quadratic formula?
Both solve the same equations; they return different things, so the choice depends on what the problem is really asking for.
Complete the square when
- You need the vertex or the maximum/minimum value
- You are sketching the parabola or writing its equation from a graph
- The equation is headed for a circle or conic section form
- Your teacher asks for the method by name, or wants the derivation
Use the formula when
- Only the roots matter and you need them fast
- The coefficients are ugly decimals with no structure to exploit
- You are checking roots for an equation you already solved another way, for example before testing them in an absolute value calculator problem with a quadratic inside
A useful habit: solve by completing the square, then verify the roots by substitution. Each root of the worked example satisfies the original equation to the precision shown, and the pair averages to 3, the axis, as it must.
What competing calculators miss
The strongest tools for this keyword give steps; the average ones give a bare vertex form. Three gaps separate this calculator from the field.
Miss 1: answers with no graph, or a graph that is not yours. Several high-ranking pages show a generic parabola illustration. Here the SVG graph is plotted from your a, b, and c on every calculation, with your vertex, axis, and roots marked, so a wrong sign in b is visible at a glance.
Miss 2: no geometric explanation. Step solvers treat "add (b/2)²" as an incantation. The area diagram shows the actual square being completed, corner and all, which is the explanation that makes the rule memorable and the errors diagnosable.
Miss 3: complex roots treated as an error. Tools that stop at "no real solutions" abandon exactly the cases algebra courses ask about next. A negative discriminant here returns the complex pair in a ± bi form with the reasoning written out, and the graph shows why no crossing exists.
One honest limit: the tool works on the equation equal to zero and reports numbers to four decimal places, trimming trailing zeros. If your course wants exact surd form such as 3 ± √5.5 left unsimplified, use the printed steps to write it yourself; the step list keeps the pre-root stage visible for that reason.
Frequently Asked Questions
It is an algebra technique that rewrites a quadratic ax^2 + bx + c as a(x - h)^2 + k. The rewritten form reveals the parabola's vertex directly and lets you solve the equation by taking a square root.
Factor a out of the x terms if it is not 1. Take half of the resulting x coefficient and square it. Add and subtract that value inside the brackets so the first three terms form a perfect square. Tidy the constants into k, then read off the vertex (h, k) or take square roots to solve.
Because the vertex form is the goal as often as the roots are. Graphing, optimisation, circle equations, and calculus all want a(x - h)^2 + k, and the quadratic formula only gives roots. Completing the square is also how the quadratic formula itself is derived.
Then the equation has two complex roots, and the parabola never crosses the x-axis. The tool shows them as x = real +/- imaginary i, and the graph shows a curve floating entirely above or below the axis.
Your actual parabola with the vertex marked, the axis of symmetry dashed, and real roots labeled at their x-intercepts, drawn from the same h, k, and roots the calculator computed.
Yes. Any non-zero a works, including decimals and negatives. A negative a flips the parabola so the vertex is a maximum instead of a minimum. Only a = 0 is refused, because that is a linear equation, not a quadratic.
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References
- Khan Academy, "Solving quadratics by completing the square" - the standard lesson sequence for the method, from move-the-constant through taking square roots. Khan Academy: Completing the square. Accessed October 2026.
- Khan Academy, "Finding the vertex of a parabola in standard form" - the vertex formula h = -b/(2a) and its derivation from the roots midpoint. Khan Academy: Vertex of a parabola. Accessed October 2026.
- Khan Academy, "Rewriting a quadratic function to find roots and vertex" - converting between standard and vertex forms and reading features from each. Khan Academy: Quadratic forms and features. Accessed October 2026.
- Khan Academy, "Worked example: quadratic formula" - the formula that completing the square derives, shown solving the same class of equations. Khan Academy: Quadratic formula worked example. Accessed October 2026.