Matemática

Quadratic Formula Calculator — Solve ax²+bx+c=0 Instantly

Solve any quadratic equation ax²+bx+c=0 with the quadratic formula. Get both roots (x₁, x₂) and the discriminant (Δ = b²−4ac) instantly — handles real, repeated, and complex roots.

🗓️ Updated June 2026 Reviewed by
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Enter a, b, and c to solve ax² + bx + c = 0 using the quadratic formula x = (−b ± √(b²−4ac)) / (2a). The calculator returns both roots and the discriminant Δ = b²−4ac, which determines the nature of the solution: Δ > 0 → two distinct real roots, Δ = 0 → one repeated real root, Δ < 0 → two complex conjugate roots. Useful for factoring parabolas, solving projectile-motion equations, break-even analysis, and any second-degree polynomial that arises in algebra, physics, or engineering.

When to use this calculator

  • Determine the two moments when a ball thrown upward at 20 m/s from a 5 m platform hits the ground: −5t² + 20t + 5 = 0.
  • Find break-even quantities for a business with revenue R(x) = −2x² + 80x − 600 by solving −2x² + 80x − 600 = 0.
  • Calculate dimensions of a rectangular garden with area 36 ft² and perimeter 26 ft, reducing to x² − 13x + 36 = 0.
  • Verify whether a conic section has real intersections by checking the discriminant sign before plotting.
  • Solve for the resonance frequency in an RLC circuit equation that simplifies to a quadratic.

Famous & Real-World Quadratic Equations (worked)

Recognizable second-degree equations from math, physics, finance and geometry — with discriminant and exact roots.

EquationabcΔ = b²−4acRoots (x₁, x₂)Where it appears
x² − x − 1 = 01−1−151.618, −0.618Golden ratio φ = (1+√5)/2
x² − 13x + 36 = 01−1336259, 4Rectangle: area 36, perimeter 26
−2x² + 80x − 600 = 0−280−6001,60010, 30Business break-even quantities
x² − 4x + 1 = 01−41123.732, 0.2682 ± √3 (irrational roots)
x² − 6x + 9 = 01−6903 (double root)Perfect-square trinomial (x−3)²
x² + x + 1 = 0111−3−½ ± i·√3/2Complex conjugate roots (Δ<0)
−4.9t² + 15t + 20 = 0−4.915206174.07 s (t>0)Projectile from 20 m at 15 m/s

Δ>0 → two real roots; Δ=0 → one repeated root; Δ<0 → two complex conjugate roots. Verify any pair with Vieta: x₁+x₂=−b/a and x₁·x₂=c/a. Projectile root keeps only the positive (physical) time.

How it works

How the Quadratic Formula Works

The quadratic formula is derived by completing the square on ax² + bx + c = 0:

Step 1 — Compute the discriminant:
  Δ = b² − 4ac

Step 2 — Apply the quadratic formula:
  x₁ = (−b + √Δ) / (2a)
  x₂ = (−b − √Δ) / (2a)

Step 3 — Interpret Δ:
  Δ > 0  →  two distinct real roots
  Δ = 0  →  one repeated real root: x = −b / (2a)
  Δ < 0  →  two complex conjugate roots: x = −b/(2a) ± i·√|Δ|/(2a)

Vieta's formulas let you double-check your answer:

  x₁ + x₂ = −b / a
  x₁ · x₂ =  c / a

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Discriminant Reference Table

Discriminant (Δ)Root typeExample equationRoots
Δ > 0 (perfect square)2 rational real rootsx² − 5x + 6 = 0 (Δ=1)x₁=3, x₂=2
Δ > 0 (non-perfect square)2 irrational real rootsx² − 4x + 1 = 0 (Δ=12)x=2±√3
Δ = 01 repeated real rootx² − 6x + 9 = 0 (Δ=0)x=3 (double)
Δ < 02 complex conjugate rootsx² + x + 1 = 0 (Δ=−3)x=−½ ± i√3/2

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Quick-Reference: Common Equations Solved

EquationabcΔx₁x₂
x²−5x+6=01−56132
x²−6x+9=01−69033 (double)
x²+x+1=0111−3−½+i√3/2−½−i√3/2
2x²−3x−2=02−3−2252−0.5
x²−4x+1=01−41122+√3≈3.7322−√3≈0.268
−x²+4x−3=0−14−3431

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Worked Examples

Example 1 — Two distinct rational roots


Equation: x² − 5x + 6 = 0 (a=1, b=−5, c=6)
  • Δ = (−5)² − 4(1)(6) = 25 − 24 = 1

  • x₁ = (5 + 1) / 2 = 3

  • x₂ = (5 − 1) / 2 = 2

  • Check (Vieta): 3+2=5=−(−5)/1 ✔ | 3×2=6=6/1 ✔
  • Example 2 — One repeated root (double root)


    Equation: x² − 6x + 9 = 0 (a=1, b=−6, c=9)
  • Δ = 36 − 36 = 0

  • x = 6 / 2 = 3 (double root — parabola is tangent to x-axis)
  • Example 3 — Complex roots


    Equation: x² + 2x + 5 = 0 (a=1, b=2, c=5)
  • Δ = 4 − 20 = −16

  • x₁ = −1 + 2i, x₂ = −1 − 2i

  • These appear in underdamped RLC circuits (no real resonance crossing).
  • Example 4 — Projectile motion


    A ball is thrown upward at 20 m/s from a 5 m height: −5t² + 20t + 5 = 0
  • a=−5, b=20, c=5

  • Δ = 400 − 4(−5)(5) = 400 + 100 = 500

  • t₁ = (−20 + √500) / (−10) ≈ −0.236 s (rejected, negative)

  • t₂ = (−20 − √500) / (−10) ≈ 4.236 s (when ball hits ground)
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    Common Mistakes to Avoid

    1. a ≠ 0 is required. If a=0 the equation is linear, not quadratic; this calculator will warn you.
    2. Sign of b in the numerator. The formula uses −b. If b=−5, then −b=+5. A sign error shifts both roots by 2b/a.
    3. 4ac includes all signs. With a=2, c=−2: 4ac = 4·2·(−2) = −16, so Δ=9−(−16)=25, not 9−16=−7.
    4. Divide the full numerator by 2a. The correct form is (−b ± √Δ) / (2a), not −b ± (√Δ / 2a).
    5. Complex roots are still solutions. When Δ<0, the equation has no real roots but always has two complex conjugate solutions in ℂ.
    6. Rearrange to standard form first. From 3x²=5x−2, rearrange to 3x²−5x+2=0 (c=+2), then identify a=3, b=−5, c=2.

    Worked Example: x² − 5x + 6 = 0

    Identify: a=1, b=−5, c=6
    Compute Δ = (−5)² − 4(1)(6) = 25 − 24 = 1
    x₁ = (5 + √1) / 2 = 6/2 = 3
    x₂ = (5 − √1) / 2 = 4/2 = 2
    Verify (Vieta): 3+2=5=−(−5)/1 ✔ | 3×2=6=6/1 ✔
    x₁=3, x₂=2 (two distinct real roots, Δ=1)

    Frequently asked questions

    What does the discriminant tell me about the graph of a parabola?
    The discriminant Δ = b²−4ac determines how the parabola y = ax²+bx+c intersects the x-axis. Δ > 0 means two x-intercepts (the parabola crosses the axis twice). Δ = 0 means the vertex just touches the x-axis at one point. Δ < 0 means the parabola floats entirely above (a>0) or below (a<0) the x-axis with no real crossings.
    How do I use the quadratic formula step by step?
    1) Write the equation in standard form ax²+bx+c=0. 2) Identify a, b, c. 3) Compute Δ = b²−4ac. 4) If Δ≥0, calculate x = (−b ± √Δ) / (2a). If Δ<0, write x = −b/(2a) ± i·√|Δ|/(2a). 5) Verify using Vieta's: x₁+x₂ = −b/a and x₁·x₂ = c/a.
    Can I use this calculator when a is negative?
    Yes. A negative leading coefficient (a < 0) produces a downward-opening parabola but the quadratic formula works identically. Example: −x²+4x−3=0 gives Δ=16−12=4, x₁=3, x₂=1. Enter a=−1, not a=1.
    Why does completing the square give the same result as the quadratic formula?
    The quadratic formula IS the result of completing the square. Starting from ax²+bx+c=0, divide by a, move c/a to the right, add (b/2a)² to both sides — you get (x+b/2a)² = Δ/(4a²), which yields x=(−b±√Δ)/(2a). The formula is just a pre-solved shortcut for completing the square.
    What are Vieta's formulas and why are they useful?
    Vieta's formulas state that x₁+x₂ = −b/a and x₁·x₂ = c/a. They let you verify roots without substituting back. For x²−5x+6=0 with roots 3 and 2: 3+2=5=−(−5)/1 ✔ and 3×2=6=6/1 ✔. They're also used to build a quadratic equation from two desired roots.
    How do I handle complex roots in practice?
    When Δ < 0, write √Δ = i·√|Δ|, giving x = −b/(2a) ± i·√|Δ|/(2a). The real part −b/(2a) is the x-coordinate of the parabola's vertex; the imaginary part measures how far the vertex is from the x-axis. In electrical engineering, complex characteristic roots correspond to oscillatory (underdamped) responses in RLC circuits.
    Is there a formula for cubic or higher-degree equations?
    Yes for cubics and quartics: Cardano's formula solves cubics (degree 3) and Ferrari's method solves quartics (degree 4). However, the Abel–Ruffini theorem (1824) proves no general radical formula exists for polynomials of degree ≥ 5. For those, numerical methods like Newton–Raphson or Durand–Kerner are used.
    What happens when b=0 or c=0?
    If b=0 and c=0, the equation is ax²=0 → double root at x=0 (Δ=0). If only c=0 (b≠0), then ax²+bx=0 factors as x(ax+b)=0, giving x=0 and x=−b/a. If only b=0 (c≠0), the equation is ax²+c=0 → x=±√(−c/a), which is real if −c/a>0 and complex otherwise.
    How does the quadratic formula apply to physics problems?
    Projectile motion gives h(t) = h₀ + v₀t − ½gt². Setting h=0 yields a quadratic where a=−½g≈−4.9 m/s² on Earth, b=v₀, c=h₀. Only the positive root is physically meaningful. Stopping distance in kinematics, lens optics, and electric potential problems also commonly reduce to quadratic equations.

    Methodology & trust

    Editorial

    Calculadora de matemática revisada por el equipo editorial de Hacé Cuentas, contrastada con NIST Digital Library of Mathematical Functions — §1.11 Algebraic and Analytic Methods, según nuestra política editorial y metodología.

    Updates

    Última revisión: June 22, 2026. Los parámetros se verifican periódicamente con las fuentes citadas.

    Privacy

    Calculations run 100% in your browser. We do not store or transmit your data.

    Limitations

    Indicative results. For critical decisions, consult a professional.

    📌 How to cite this calculator

    Rodríguez, M. (2026). Quadratic Formula Calculator — Solve ax²+bx+c=0 Instantly. Hacé Cuentas. https://hacecuentas.com/quadratic-equation-roots-discriminant

    Contenido bajo licencia CC-BY 4.0 — reutilizable citando la fuente con enlace a Hacé Cuentas.

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