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Quadratic Equation Solver

Find roots of ax² + bx + c = 0 with visual graph

ax² + bx + c = 0

Solution

Root 1 (x₁)—
Root 2 (x₂)—
Discriminant (D)—
Nature of Roots—
Vertex—
Axis of Symmetry—

The Quadratic Formula

x = (-b ± √(b² - 4ac)) / 2a

The discriminant D = b² - 4ac determines the nature of roots: D > 0 gives two distinct real roots, D = 0 gives one repeated real root, D < 0 gives two complex conjugate roots.

The quadratic formula, and what the discriminant tells you

For ax² + bx + c = 0 the roots are x = (−b ± √(b² − 4ac)) ÷ 2a. The part under the root, b² − 4ac, is the discriminant, and its sign alone tells you whether you get two real roots, one repeated root, or a complex pair — before you finish the arithmetic.

What the discriminant decides

Discriminant D = b² − 4acRootsGraph of y = ax² + bx + c
D > 0, a perfect squareTwo distinct rational rootsCrosses the x-axis twice at nice values
D > 0, not a perfect squareTwo distinct irrational rootsCrosses the x-axis twice
D = 0One repeated real root, x = −b/2aTouches the x-axis at the vertex
D < 0Two complex conjugate rootsNever touches the x-axis

Three worked examples

D > 0 — x² − 5x + 6 = 0

a = 1, b = −5, c = 6

D = 25 − 24 = 1

x = (5 ± 1) / 2

x = 3 and x = 2

D = 0 — x² + 6x + 9 = 0

a = 1, b = 6, c = 9

D = 36 − 36 = 0

x = −6 / 2

x = −3, a repeated root; the parabola just touches the axis

D < 0 — x² + x + 1 = 0

a = 1, b = 1, c = 1

D = 1 − 4 = −3

x = (−1 ± i√3) / 2

x ≈ −0.5 + 0.866i and −0.5 − 0.866i

Everything else you can read off the coefficients

When factoring beats the formula

If the discriminant is a perfect square, the equation factors over the integers and factoring is faster. For x² − 5x + 6 you need two numbers multiplying to 6 and adding to −5 — −2 and −3 — so it is (x − 2)(x − 3) on sight. When a ≠ 1, look for factors of a×c that add to b. The formula never fails, but it is slower, and on a written exam it is where sign slips creep in.

Completing the square, which is where the formula comes from

  1. Divide through by a: x² + (b/a)x + c/a = 0.
  2. Move the constant: x² + (b/a)x = −c/a.
  3. Add (b/2a)² to both sides to make the left side a perfect square.
  4. It becomes (x + b/2a)² = (b² − 4ac) / 4a².
  5. Take the square root of both sides and subtract b/2a. That is the quadratic formula.

More questions people ask

What is the quadratic formula?

For any equation of the form ax² + bx + c = 0 with a not zero, x equals minus b, plus or minus the square root of b squared minus 4ac, all divided by 2a.

What does it mean if the discriminant is negative?

The equation has no real solutions. The parabola never crosses the x-axis, and the two roots are complex conjugates of the form p + qi and p − qi.

Can a quadratic have only one solution?

Yes, when the discriminant is exactly zero. It is called a repeated or double root and sits at x = −b/2a, which is also the vertex of the parabola.

How do I find the vertex of a parabola?

The x-coordinate is −b/2a. Substitute that back into the equation to get the y-coordinate, or use y = c − b²/4a directly.

What if a equals zero?

Then it is not a quadratic at all — it collapses to the linear equation bx + c = 0 with the single solution x = −c/b. The formula fails because it would divide by zero.

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Frequently Asked Questions

What is the quadratic formula?
x = (-b ± sqrt(b²-4ac)) / 2a. It gives the roots of any quadratic equation ax²+bx+c=0.
What is the discriminant?
D = b²-4ac. If D>0: two real roots, D=0: one repeated root, D<0: two complex conjugate roots.
What is the vertex of a parabola?
The vertex is at x=-b/(2a). It's the minimum point if a>0 or maximum if a<0.

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