Completing the Square Calculator — Vertex Form (Free)
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A completing the square calculator rewrites ax² + bx + c as a(x − h)² + k. For 2x² + 8x − 10, factor out 2 to get 2(x² + 4x) − 10, then add and subtract (4/2)² = 4 inside the bracket: 2(x + 2)² − 18. The vertex is (−2, −18), and 2x² + 8x − 10 = 0 when x = 1 or x = −5.
2x² + 8x − 10 in vertex form is 2(x + 2)² − 18. The vertex is (−2, −18), and 2x² + 8x − 10 = 0 when x = −5 or x = 1.
Show steps
Vertex form by completing the square
- Factor a = 2 out of the x2 and x terms: 2(x2 + 4x) − 10.
- Take half of the x coefficient 4: 2. Square it: (2)2 = 4.
- Add and subtract 4 inside the bracket: 2(x2 + 4x + 4 − 4) − 10.
- The first three terms form a perfect square, x2 + 4x + 4 = (x + 2)2: 2(x + 2)2 − 2·4 − 10.
- Combine the constants: −8 + (−10) = −18. Vertex form: 2(x + 2)2 − 18.
- Compare with a(x − h)2 + k: h = −2 and k = −18. The vertex is (−2, −18) and the axis of symmetry is x = −2. Since a = 2 is positive, the parabola opens up and −18 is the minimum value.
Solve 2x² + 8x − 10 = 0
- Set 2x2 + 8x − 10 = 0. Move the constant term to the right: 2x2 + 8x = 10.
- Divide every term by a = 2: x2 + 4x = 5.
- Add (half of 4)2 = (2)2 = 4 to both sides: x2 + 4x + 4 = 5 + 4 = 9.
- Write the left side as a square: (x + 2)2 = 9.
- Take the square root of both sides, keeping both signs: x + 2 = ±√9 = ±3.
- Subtract 2: x = −2 ± 3, so x = 1 or x = −5.
Need the roots with the quadratic formula too? Try the quadratic equation solver. For focus and directrix, use the parabola calculator.
How to use the completing the square calculator
- Enter the coefficients a, b and c of ax² + bx + c, or open Type it and write an expression such as 2x^2 − 3x + 1 or an equation such as x^2 + 6x = 7. Fractions and decimals are fine.
- Read the vertex form with exact fractions, the vertex, the axis of symmetry, the minimum or maximum value and the solutions (real or complex).
- Look at the area model, which shows the corner square you add, and at the graph of the parabola with its vertex and x-intercepts.
- Open Show steps. You get two sets of steps: vertex form by factoring out a, and solving the equation by dividing by a.
The completing the square formula
The key identity is x² + bx + (b/2)² = (x + b/2)². Half of the x-coefficient, squared, is the number that turns x² + bx into a perfect square. For x² + 6x, half of 6 is 3 and 3² = 9, so x² + 6x + 9 = (x + 3)².
For any quadratic, completing the square gives the vertex form ax² + bx + c = a(x − h)² + k, with h = −b/(2a) and k = c − b²/(4a). The point (h, k) is the vertex of the parabola y = ax² + bx + c.
How to complete the square, step by step
To solve ax² + bx + c = 0:
- Move the constant to the right side: ax² + bx = −c.
- If a ≠ 1, divide every term by a.
- Take half of the x-coefficient and square it. Add that number to both sides.
- Write the left side as a square, (x + b/(2a))².
- Take the square root of both sides, remembering ±, and solve for x.
To get vertex form instead, do not divide by a. Factor a out of the x² and x terms, add and subtract the square inside the bracket, then multiply back out. Dividing would change the expression; factoring keeps it equal.
Worked examples
Example 1: a = 1
Solve x² + 6x = 7. Half of 6 is 3 and 3² = 9, so add 9 to both sides: x² + 6x + 9 = 16, which is (x + 3)² = 16. Then x + 3 = ±4, so x = 1 or x = −7.
Example 2: a ≠ 1 with fractions
Write 2x² + 5x − 3 in vertex form. Factor out 2: 2(x² + (5/2)x) − 3. Half of 5/2 is 5/4, and (5/4)² = 25/16. So 2(x² + (5/2)x + 25/16 − 25/16) − 3 = 2(x + 5/4)² − 25/8 − 3 = 2(x + 5/4)² − 49/8. The vertex is (−5/4, −49/8). Setting it to 0: (x + 5/4)² = 49/16, so x + 5/4 = ±7/4, giving x = 1/2 or x = −3.
Example 3: irrational roots
Solve 3x² − 5x + 1 = 0. Divide by 3: x² − (5/3)x = −1/3. Half of −5/3 is −5/6, and (−5/6)² = 25/36. Adding it gives (x − 5/6)² = −1/3 + 25/36 = 13/36. So x = 5/6 ± √13/6 = (5 ± √13)/6, about 1.4343 and 0.2324.
Example 4: complex roots
Solve x² + 4x + 13 = 0. Move 13: x² + 4x = −13. Add 4: (x + 2)² = −9. No real number squared is negative, so there are no real solutions. With √(−1) = i, x + 2 = ±3i and x = −2 ± 3i.
Why it is called completing the square
Picture x² as a square with side x, and bx as two rectangles, each x by b/2, placed along two sides of it. The shape is almost a larger square with side x + b/2, but one corner is missing. That corner is a (b/2) by (b/2) square, area (b/2)². Adding it “completes the square”. The calculator draws this area model for your numbers.
| Expression | Add (b/2)² | Perfect square |
|---|---|---|
| x² + 2x | 1 | (x + 1)² |
| x² + 4x | 4 | (x + 2)² |
| x² − 6x | 9 | (x − 3)² |
| x² + 10x | 25 | (x + 5)² |
| x² − 3x | 9/4 | (x − 3/2)² |
| x² + x | 1/4 | (x + 1/2)² |
Where completing the square is used
- Graphing parabolas. Vertex form shows the vertex, the axis of symmetry and the minimum or maximum at a glance. The parabola calculator adds the focus and directrix.
- The quadratic formula. Completing the square on ax² + bx + c = 0 gives x = (−b ± √(b² − 4ac)) / (2a). The quadratic equation solver compares both methods.
- Circles. Completing the square in x and in y turns x² + y² − 6y = 16 into x² + (y − 3)² = 25, a circle with center (0, 3) and radius 5. The calculator handles this too, and the circle equation calculator goes further.
Common mistakes
- Adding the square to one side only. When solving, add (b/2)² to both sides. When writing vertex form, add and subtract it on the same side.
- Forgetting to factor out a. For 2x² + 8x, half of 8 is not the number to use; factor first, 2(x² + 4x), and halve 4.
- Not multiplying the subtracted square by a. In 2(x² + 4x + 4 − 4), the −4 becomes −8 when it leaves the bracket.
- Dropping the ±. (x + 3)² = 16 has two solutions, x = 1 and x = −7.
- Sign of h. (x + 2)² means h = −2, so the vertex is to the left of the y-axis.
To check your work, expand your vertex form. 2(x + 2)² − 18 = 2(x² + 4x + 4) − 18 = 2x² + 8x − 10, which matches. You can also substitute a root back into the original equation.
Completing the square practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
Half of −8 is −4 and (−4)² = 16: x² − 8x + 16 − 16 + 3 = (x − 4)² − 13.
Factor 2: 2(x² + 6x) + 7. Add and subtract 9 inside: 2(x + 3)² − 18 + 7 = 2(x + 3)² − 11.
After 3x² + 6x = 9, divide by 3 so the x² coefficient is 1: x² + 2x = 3. Then add (2/2)² = 1.
−(x² − 4x) + 1 = −(x − 2)² + 4 + 1 = −(x − 2)² + 5. Since a is negative the vertex (2, 5) is a maximum.
Complete the square in y: y² − 6y + 9 = (y − 3)², so x² + (y − 3)² = 16 + 9 = 25 and r = 5.
Quiz complete
Frequently asked questions
What is completing the square?
It is a way of rewriting a quadratic so it contains a perfect square, such as x² + 6x + 5 = (x + 3)² − 4. It is used to solve quadratic equations, to find the vertex of a parabola and to put circle equations in standard form.
How do you solve a quadratic by completing the square?
Move the constant to the right, divide by a, add (half the x-coefficient)² to both sides, write the left side as a square, then take square roots with ±. For x² + 6x = 7: (x + 3)² = 16, so x = 1 or x = −7.
What is the equation for completing the square?
x² + bx + (b/2)² = (x + b/2)². For a general quadratic, ax² + bx + c = a(x − h)² + k with h = −b/(2a) and k = c − b²/(4a).
Is completing the square the same as the quadratic formula?
They give the same answers, because the quadratic formula comes from completing the square on ax² + bx + c = 0. Completing the square also gives vertex form; the formula is quicker when you only need the roots.
Is completing the square a non-calculator topic?
Usually, yes. Exams often expect it by hand, and the numbers are chosen to work out neatly. Practice the steps by hand, and use the calculator to check your vertex form and your solutions.
How do you use completing the square to get vertex form?
Factor a out of the x² and x terms, add and subtract (b/(2a))² inside the bracket, write the perfect square, and combine the constants. For 2x² + 8x − 10, you get 2(x + 2)² − 18, so the vertex is (−2, −18).
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