Enter a, b, c for ax²+bx+c and get the vertex form a(x−h)²+k with every algebraic step shown, plus the vertex, discriminant, roots and a parabola graph.
Vertex form
Vertex (h, k)
Axis of symmetry
Discriminant D = b²−4ac
Roots
y-intercept
Opens
Step-by-step
Parabola graph
Frequently asked questions
What does "completing the square" mean?
It is a technique that rewrites a quadratic ax^2+bx+c in vertex form a(x-h)^2+k. You force the x-terms into a perfect square trinomial by adding and subtracting (b/2a)^2, which reveals the vertex (h, k) directly. For example, x^2+6x+5 becomes (x+3)^2-4, so the vertex is (-3, -4).
What is the formula for completing the square?
For ax^2+bx+c with a not equal to 0, the vertex form is a(x-h)^2+k where h = -b/(2a) and k = c - b^2/(4a). Equivalently, k = f(h), the function value at x = h. For x^2-5x+6: h = 5/2, k = -1/4, so the vertex form is (x-5/2)^2-1/4.
Can you walk through an example step by step?
Take x^2+6x+5. Step 1: x^2+6x+5. Step 2: factor a = 1 out of the x-terms: (x^2+6x)+5. Step 3: half of 6 is 3, and 3^2 = 9. Step 4: add and subtract 9: (x^2+6x+9)-9+5. Step 5: factor the perfect square: (x+3)^2-9+5. Step 6: combine constants: (x+3)^2-4. The vertex is (-3, -4).
What if a is not 1?
Factor a out of the x-terms first, then complete the square inside the parentheses. Example: 2x^2+8x+3 = 2(x^2+4x)+3 = 2(x^2+4x+4)-8+3 = 2(x+2)^2-5. The vertex is (-2, -5).
How do I find the vertex from the completed square?
In a(x-h)^2+k the vertex is (h, k) - read it straight off. Watch the sign: (x+3)^2-4 means h = -3 and k = -4, so the vertex is (-3, -4). If a is greater than 0 the vertex is the minimum point; if a is less than 0 it is the maximum point.
What is the discriminant, and how does it relate to completing the square?
The discriminant is D = b^2-4ac. From vertex form, the roots are x = h +/- sqrt(D)/(2a). For x^2-5x+6: h = 5/2 and D = 1, giving x = 5/2 +/- 1/2, that is x = 2 and x = 3. If D is greater than 0 there are two real roots, if D = 0 one repeated root, if D is less than 0 no real roots.
What happens when the discriminant is negative?
There are no real roots, but completing the square still works and the vertex is still real. Example: x^2+x+1 = (x+1/2)^2+3/4, vertex (-1/2, 3/4), D = -3. The parabola never crosses the x-axis; its minimum value is 3/4.
When is completing the square better than the quadratic formula?
Use completing the square when you need the vertex, the maximum or minimum value, or the shape of the graph - the formula only gives the roots. In fact, the quadratic formula itself is derived by completing the square on ax^2+bx+c = 0.
Can I complete the square if a = 0?
No. If a = 0 the expression is linear (bx+c), not quadratic, and the formulas divide by a. This calculator rejects a = 0 and explains why.
How is this different from the Quadratic Equation Solver?
This tool is a conversion specialist: it turns standard form into vertex form with full pedagogical steps, plus the vertex, discriminant, and graph. If your goal is only to find x (the roots), the Quadratic Equation Solver finds them through several methods including the quadratic formula, factoring, and completing the square.
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