⬜ Completing the Square Calculator
2x² − 4x + 5 = 2(x − 1)² + 3 — completed exactly, with the halve-and-square working written out, plus the vertex (1, 3) it hands you for free.
Coefficients accept integers, decimals and fractions. Need the roots too? Use the quadratic solver.
2(x − 1)² + 3
(1, 3)
vertex (h, k)
x = 1
axis of symmetry
minimum 3
extreme value
Working (completing the square)
- Factor a out of the x-terms: 2(x² − 2x) + 5
- Take half of the x-coefficient, -2 ÷ 2 = -1, and square it: 1
- Add and subtract it inside, regroup: 2(x − 1)² − 2·1 + 5
- Collect the constants: k = 5 − 2 = 3 → 2(x − 1)² + 3
Vertex form with exact fractions — 2x² − 4x + 5 becomes 2(x − 1)² + 3, every constant exact. Runs locally.
How the completing the square calculator works
Completing the square rewrites a quadratic in vertex form: factor a out of the x-terms, take half the x-coefficient and square it, add and subtract that square inside, regroup. The calculator performs each of those steps in exact rational arithmetic and displays them numbered — so 3x² + 5x − 1 comes out as 3(x + 5/6)² − 37/12 with honest fractions, where decimal calculators print 3(x + 0.833…)² − 3.083… and the exactness is gone. The vertex (−h, k), axis of symmetry and minimum/maximum value are read straight off the result.
Completing the square is the technique — it's how the quadratic formula is derived, how circles are extracted from x² + y² + Dx + Ey + F = 0, and how integrals with quadratics get solved. The vertex form answers a different question than the roots: where is the extreme point? For the roots themselves, the quadratic solver is the right tool, and the two pages cross-link.
Frequently asked questions
What are the steps to complete the square?
Factor a from the x-terms; take half of the x-coefficient and square it; add and subtract that square; regroup into a(x + h)² + k. The calculator shows each step with your exact numbers.
What is vertex form good for?
It exposes the parabola's vertex (−h, k), axis of symmetry and minimum or maximum instantly — the form for optimization problems and graphing. Standard form hides all three.
How does this relate to the quadratic formula?
The formula IS completing the square done once in general: apply the steps to ax² + bx + c = 0 symbolically and x = (−b ± √(b² − 4ac))/2a falls out. Doing it numerically here is the same walk.
What if a isn't 1?
Factor it out of the x-terms first — the step decimal calculators fumble. 2x² − 4x + 5 → 2(x² − 2x) + 5 → 2(x − 1)² + 3, kept exact throughout.
Does it solve for the roots too?
That's the quadratic equation solver's job (linked on this page) — it gives exact roots, the discriminant and the factored form. This page gives the vertex-form identity of the same quadratic.