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Quadratic Equation quadratic formula.

Solve ax² + bx + c = 0 – see the discriminant and one or two real solutions instantly

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Quadratic Equation

Solve ax² + bx + c = 0 with the quadratic formula

1x² − 5x + 6 = 0
x₁

3

First solution

x₂

2

Second solution

Discriminant D = 1

Quadratic formula:

x = (−b ± √D) / (2a)

D = b² − 4ac

The quadratic formula – step by step

Understand and solve quadratic equations with confidence

The quadratic equation ax² + bx + c = 0 is one of the most frequently asked topics in school mathematics. It arises whenever parabolas cross the x-axis, in optimization problems (minimum/maximum of a quadratic function), in physics (projectile motion, lever laws) and in engineering mathematics. The quadratic formula x = (−b ± √D) / (2a) always gives the result – provided the discriminant D is not negative.

The discriminant D = b² − 4ac is the decisive value. It sits under the square root and determines whether the equation has two (D > 0), one (D = 0) or no real solution (D < 0). If D < 0, the square root of a negative number would be required, which is impossible within the real numbers. The corresponding parabola f(x) = ax² + bx + c then does not cross the x-axis.

For simple equations with integer solutions, the factoring method is also efficient: for x² − 5x + 6 = 0, you look for two numbers with a sum of −5 and a product of +6. These are −3 and −2, so (x − 3)(x − 2) = 0 → x = 3 or x = 2. This method works without a calculator, but only when integer solutions exist. The quadratic formula always works – even for irrational solutions like x = (1 + √5) / 2 (the golden ratio).

In practice, quadratic equations appear in motion problems (solving s = v₀t + ½at² for t), area optimization (maximum area of a rectangle with a given perimeter) and circle geometry. Our calculator solves ax² + bx + c = 0 for any real coefficients a, b, c and shows the discriminant as well as all real solutions with six decimal places of precision.

Calculation example: x² − 5x + 6 = 0

Step 1: calculate the discriminant

Step 1: calculate the discriminant
ItemAmount
Coefficientsa = 1, b = −5, c = 6
D = b² − 4ac(−5)² − 4×1×6
D = 25 − 241

Step 2: calculate the solutions (D = 1 > 0)

Step 2: calculate the solutions (D = 1 > 0)
ItemAmount
x₁ = (−b + √D) / (2a) = (5 + 1) / 23
x₂ = (−b − √D) / (2a) = (5 − 1) / 22

Frequently asked questions about the quadratic equation

Quadratic formula, discriminant and solving strategies

The quadratic formula solves any quadratic equation ax² + bx + c = 0: x₁,₂ = (−b ± √(b² − 4ac)) / (2a). The plus sign gives x₁, the minus sign gives x₂. Requirement: a ≠ 0 (otherwise the equation is linear, not quadratic).

The discriminant D = b² − 4ac determines the number of real solutions: D > 0: two distinct real solutions (x₁ ≠ x₂). D = 0: exactly one solution (a double root, x₁ = x₂ = −b/(2a)). D < 0: no real solution (the parabola does not cross the x-axis). The discriminant is the first thing you should check for any quadratic equation.

Coefficients: a=1, b=−5, c=6. Discriminant: D = (−5)² − 4×1×6 = 25 − 24 = 1. Since D = 1 > 0, there are two solutions: x₁ = (5 + √1) / 2 = (5+1)/2 = 3 and x₂ = (5 − 1) / 2 = 2. Check: (x−3)(x−2) = x² − 5x + 6 ✓. The solutions are x₁ = 3 and x₂ = 2.

A double root occurs when the discriminant D = 0. In this case, the corresponding parabola touches the x-axis at exactly one point (a tangent point) instead of crossing it. The formula simplifies to x = −b / (2a). Example: x² − 2x + 1 = 0 has D = 4 − 4 = 0, so a double root x = 1 (the parabola y = (x−1)² touches the x-axis at x=1).

For "nice" solutions, you can factor the equation: which two numbers add up to the coefficient b and multiply to the coefficient c? For x² − 5x + 6: you need numbers with a sum of −5 and a product of 6: these are −3 and −2. So (x−3)(x−2) = 0 → x=3 or x=2. For more complicated coefficients, the quadratic formula is more efficient.

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