Completing the Square, Step by Step: From ax² + bx + c to Vertex Form

Every quadratic hides a parabola, and every parabola has one special point — the vertex — where it turns around. The trouble is that the standard form ax² + bx + c keeps that point buried. Completing the square is the algebra that digs it out. It rewrites the quadratic as a(x − h)² + k, a shape that hands you the vertex (h, k) and the axis of symmetry x = h with nothing left to compute.

This is the method behind the completing the square calculator, and it's worth understanding by hand, because once the steps click, you can find a vertex, solve an equation, and graph a parabola without reaching for a formula you half-remember.

The one move that makes it work

The whole method rests on a single algebraic fact: x² + bx becomes a perfect square if you add and subtract the square of half the linear coefficient. Take half of b, square it, and you get the missing piece — (b/2)² — that turns x² + bx into (x + b/2)². You add it to complete the square and subtract it again so the expression stays equal to what you started with.

That "add it, then subtract it" balance is the part people skip, and skipping it quietly changes the answer. You aren't allowed to add 9 to an expression for free; you add 9 and remove 9 in the same breath, then regroup.

A worked example: x² + 6x + 5

Let's run the full procedure on x² + 6x + 5.

Look at the linear coefficient. Here b = 6. Half of 6 is 3, and 3² = 9. That 9 is the number we'll add and subtract.
Add and subtract 9. Rewrite x² + 6x + 5 as x² + 6x + 9 − 9 + 5. Nothing has changed; we've added zero.
Fold the perfect square. The first three terms, x² + 6x + 9, are exactly (x + 3)². So the expression becomes (x + 3)² − 9 + 5.
Collect the constant. −9 + 5 = −4. The vertex form is (x + 3)² − 4.

Read it off: h = −3 (note the sign flips — the bracket says +3, the vertex sits at −3), k = −4. The vertex is (−3, −4), the axis of symmetry is x = −3, and since the coefficient on the squared term is positive, the parabola opens upward and −4 is its minimum value.

That's the entire story in four lines. The calculator shows these same lines numbered, so you can copy the method onto a test paper where the rubric grades the work, not just the answer.

When a isn't 1: factor first

If the leading coefficient isn't 1, there's one extra move at the front, and it's the step most often botched. Factor a out of the x-terms before you halve anything.

Take 2x² + 8x + 3. Pull the 2 out of the first two terms: 2(x² + 4x) + 3. Now you complete the square inside the bracket on x² + 4x. Half of 4 is 2, 2² = 4, so add and subtract 4 inside: 2(x² + 4x + 4 − 4) + 3. The inner perfect square is (x + 2)², leaving 2((x + 2)² − 4) + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5.

The vertex is (−2, −5). If you'd halved the 8 instead of the 4 — that is, forgotten to factor the 2 out first — you'd land on a wrong inner term and a broken vertex. The shortcut formulas h = −b/(2a) and k = c − b²/(4a) encode exactly this work: for 2x² + 8x + 3, h = −8/4 = −2 and k = 3 − 64/8 = −5. Same answer, less writing, but the derivation is what tells you why.

Solving the equation, not just finding the vertex

Vertex form does more than locate the turning point — it solves the quadratic directly, without the formula. Once you have a(x − h)² + k = 0, isolate the square and take a root.

Back to x² + 6x + 5 = (x + 3)² − 4. Set it to zero: (x + 3)² − 4 = 0, so (x + 3)² = 4. Take the square root of both sides: x + 3 = ±2. That gives x = −3 + 2 = −1 and x = −3 − 2 = −5. Two real roots, no formula memorized.

The general pattern is x = h ± √(−k/a). The sign of −k/a tells you the whole story: positive gives two real roots, zero gives one repeated root, and negative gives a conjugate pair. For x² + 2x + 5 = (x + 1)² + 4, you'd get (x + 1)² = −4, so x = −1 ± 2i. When you only want the numeric roots and don't care about the shape, the quadratic equation solver is the faster route; completing the square earns its keep when the vertex and the min/max matter too.

Why this is also where the quadratic formula comes from

Here's the part that made it finally make sense for me. I'd treated the quadratic formula as a magic incantation for years — plug in a, b, c, turn the crank. Then a teacher completed the square on the general quadratic ax² + bx + c = 0 on the board, factored a out, added and subtracted (b/2a)², folded the square, and took the root. Out fell x = (−b ± √(b² − 4ac)) / 2a, line by line. The formula wasn't handed down from somewhere; it's just completing the square done once on letters instead of numbers. The discriminant b² − 4ac is exactly the thing under the square root that decides real-versus-complex. After seeing that derivation, the formula stopped being something I memorized and became something I could rebuild from scratch if I ever forgot it.

Reading the parabola off vertex form

Once a quadratic is in a(x − h)² + k, every graph-worthy fact is sitting in plain sight:

Vertex: (h, k), straight from the form.
Axis of symmetry: the vertical line x = h, which always passes through the vertex.
Direction: opens up when a > 0, down when a < 0.
Min or max: k is the minimum when a > 0 and the maximum when a < 0.

For a model like −x² + 4x − 1, completing the square gives −(x − 2)² + 3. Because a < 0 the parabola opens downward, so k = 3 is the maximum, reached at x = 2 — no calculus needed to find the peak. That's why completing the square shows up in optimization problems long before derivatives do.

Try a few coefficients of your own in the completing the square calculator: type a, b, and c, and it factors, halves, folds, and solves with every step visible. The a/b/c values live in the URL, so a shared link reopens the exact same worked example — handy for a homework hint or a lesson slide. Work a couple by hand first, then check yourself against the tool, and the method stops feeling like a trick and starts feeling like the obvious thing to do.

Made by Toolora · Updated 2026-06-13