Case 1
- Coefficient a (before x²)
- 1
- Coefficient b (before x)
- -5
- Constant c
- 6
x₁ = 3, x₂ = 2
Open with these valuesQuadratic Formula
x₁ = 3, x₂ = 2
Result: x₁ = 3, x₂ = 2Enter a, b and c. The discriminant D = b² − 4ac decides everything: D above zero gives two real roots, D = 0 one repeated root, D below zero a pair of complex conjugates. For x² − 5x + 6 = 0 the roots are x₁ = 3 and x₂ = 2.
The numbers at a glance
Held fixed: Coefficient a (before x²) 1.0000, Coefficient b (before x) -5.0000.
| Constant c | Result |
|---|---|
| 0.0000 | 5.000000 |
| 2.0000 | 4.561553 |
| 4.0000 | 4.000000 |
| 6.0000Your value | 3.000000 |
| 8.0000 | 2.500000 |
| 10.0000 | 2.500000 |
| 12.0000 | 2.500000 |
Worked examples
Case 2
- Coefficient a (before x²)
- 2
- Coefficient b (before x)
- -7
- Constant c
- 3
x₁ = 3, x₂ = 0.5
Open with these valuesCase 3
- Coefficient a (before x²)
- 1
- Coefficient b (before x)
- 2
- Constant c
- 5
x₁ = −1 + 2i, x₂ = −1 − 2i (no real root)
Open with these valuesHow it's calculated
x = (−b ± √(b² − 4ac)) ÷ 2a
- StepWrite the equation as ax² + bx + c = 0 and enter all three coefficients with their signs.
- StepCompute the discriminant D = b² − 4ac — its sign decides how many real roots exist.
- ResultApply both signs: x = (−b + √D) ÷ 2a and x = (−b − √D) ÷ 2a.
What this number means
The quadratic formula solves every equation of the form ax² + bx + c = 0 in which a is not zero. It returns the values of x that make the expression zero — the points where the parabola meets the x-axis. Everything hinges on the discriminant D = b² − 4ac, the part under the square root. A positive D means the parabola crosses the axis twice, so there are two distinct real roots. A D of exactly zero means it only touches the axis, and the two roots collapse into a single repeated value. A negative D means the parabola never reaches the axis: there is no real root, and the two solutions are a pair of complex conjugates. This page still answers in that case and reports the pair as p ± qi, where p = −b ÷ 2a is the real part both share. The large number above is the larger of the two real roots — or that shared real part when no real root exists. Because the formula works for every quadratic, it succeeds where factoring stalls: awkward decimals, large numbers and equations without neat factors all yield to it at once.
a must not be zero
With a = 0 there is no x² term and the equation is linear, not quadratic. The page then solves bx + c = 0 instead and says so; with a = 0 and b = 0 nothing is left to solve.
A negative discriminant is still an answer
When D is below zero the two solutions are complex conjugates p ± qi. The page prints both and marks that there is no real root.
Check by substituting
Put each root back into ax² + bx + c and the result should be zero. A second check: the two roots add up to −b ÷ a and multiply to c ÷ a.
Commonly misread
A negative discriminant means the equation has no solution.
It has no REAL solution. The two solutions are complex conjugates, shown here as p ± qi.
For x² − 5x + 6 = 0 the coefficient b is 5.
It is −5 — the sign belongs to the coefficient. A misplaced minus changes both roots completely.
The larger root is always (−b + √D) ÷ 2a.
Only while a is positive. With a negative a that expression is the smaller root.
Reference table
| a, b, c | D = b² − 4ac | Roots |
|---|---|---|
| 1, −5, 6 | 1 | x₁ = 3, x₂ = 2 |
| 1, −4, 4 | 0 | x₁ = x₂ = 2 |
| 1, 0, −9 | 36 | x₁ = 3, x₂ = −3 |
| 2, −7, 3 | 25 | x₁ = 3, x₂ = 0.5 |
| 1, 2, 5 | −16 | −1 ± 2i (no real root) |
Questions
How do I use the quadratic formula?
Write your equation as ax² + bx + c = 0, then enter the three coefficients a, b and c. The formula x = (−b ± √(b² − 4ac)) ÷ 2a produces both roots — for x² − 5x + 6 = 0 that is x₁ = 3 and x₂ = 2.
What is the discriminant and what does it tell me?
The discriminant is the part under the square root, D = b² − 4ac. Its sign decides how many real roots exist: above zero gives two distinct real roots, exactly zero gives one repeated root, below zero gives none.
What happens when there are no real roots?
The square root of a negative number has no real value, so the parabola never crosses the x-axis. The two solutions are then complex conjugates, and this page reports them as p ± qi — for x² + 2x + 5 = 0 that is −1 ± 2i.
Why can't coefficient a be zero?
Without an x² term the equation is linear, bx + c = 0, and the formula would divide by zero in its 2a denominator. This page falls back to the single linear solution x = −c ÷ b and labels it as such.
Can I enter negative coefficients?
Yes, any of a, b and c may be negative, positive or — for b and c — zero. For 2x² − 7x + 3 = 0 you enter a = 2, b = −7, c = 3, which gives x₁ = 3 and x₂ = 0.5.
Sources and last check
Information, not professional advice.