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The Quadratic Formula over the Complex Numbers

The quadratic formula, x=b±b24ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, solves every quadratic ax2+bx+c=0ax^2 + bx + c = 0 — even ones that do not factor.

The piece under the root, the discriminant b24acb^2 - 4ac, previews the answer: positive gives two real solutions, zero gives one, and negative gives a complex pair.

Applying the formula

Identify aa, bb, and cc from standard form, then substitute into the formula and simplify. Take care with signs, especially a negative bb.

The ±\pm produces the two solutions. Reduce the fraction and the radical at the end.

The discriminant

The discriminant b24acb^2 - 4ac decides the nature of the roots. Positive means two distinct real solutions; zero means one repeated real solution.

Negative means the root is imaginary, so the two solutions are complex conjugates like 2±3i2 \pm 3i. Checking the discriminant first tells you what kind of answer to expect.

Worked examples

Example 1: two real solutions

Solve 2x25x3=02x^2 - 5x - 3 = 0 with the quadratic formula.

Identify a, b, ca=2,b=5,c=3a = 2, b = -5, c = -3
Substitutex=5±25+244x = \dfrac{5 \pm \sqrt{25 + 24}}{4}
Simplify the rootx=5±74x = \dfrac{5 \pm 7}{4}

Answer: x=3x = 3 or x=12x = -\dfrac{1}{2}

Example 2: a complex pair

What does a discriminant of 16-16 tell you?

Negative discriminantb24ac<0b^2 - 4ac < 0
Root of a negative is imaginary16=4i\sqrt{-16} = 4i

Answer: Two complex-conjugate solutions

Example 3: a discriminant of zero

Solve x26x+9=0x^2 - 6x + 9 = 0 with the quadratic formula.

Identify a, b, ca=1,b=6,c=9a = 1, b = -6, c = 9
Compute the discriminant3636=036 - 36 = 0
The root is zero, so both signs give the same valuex=62x = \dfrac{6}{2}

Answer: x=3x = 3, one repeated real solution

Try one yourself

Common questions

When should I use the quadratic formula?

Any time a quadratic is hard to factor, or when you need exact solutions. It works on every quadratic in standard form.

What does the discriminant tell me?

b24acb^2 - 4ac reveals the solution type: positive → two real, zero → one real (repeated), negative → two complex.

How do complex solutions appear?

When the discriminant is negative, the square root produces ii, giving a conjugate pair like 2±3i2 \pm 3i.

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