Solve ax² + bx + c = 0 with the quadratic formula, including repeated roots and complex conjugate pairs.
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For x² − 5x + 6 = 0, entering a = 1, b = −5 and c = 6 gives the roots 3 and 2 with a discriminant of 1. For 2x² − 4x − 6 = 0 the roots are 3 and −1, and x² + 2x + 5 = 0 gives x = −1 ± 2i.
The discriminant b² − 4ac is calculated first. When it is positive there are two real roots, (−b ± √discriminant) ÷ (2a); when it is 0 both branches give the same repeated root; and when it is negative the tool returns a complex conjugate pair as a real part plus or minus an imaginary part. The coefficient a cannot be 0, because the equation would then be linear.
Rewrite the equation in standard form before entering the values, moving every term to one side and keeping the signs. A missing term has a coefficient of 0, as in x² − 2 = 0 where b = 0. Irrational roots are displayed rounded to eight decimal places, so x² − 2 = 0 shows ±1.41421356.
Rewrite the equation as ax² + bx + c = 0 before entering values, moving every term to one side and keeping the signs. For x² − 5x + 6 = 0, enter a = 1, b = −5 and c = 6; the roots are 3 and 2. For 2x² = 4x + 6, subtract to get 2x² − 4x − 6 = 0, so a = 2, b = −4 and c = −6, and the roots are 3 and −1. A missing term has a coefficient of 0, as in x² − 2 = 0 where b = 0.
The discriminant b² − 4ac decides the kind of answer before any square root is taken. A positive value gives two different real roots: x² − 2 = 0 has a discriminant of 8 and roots of about ±1.41421356. Zero gives one repeated root: x² + 2x + 1 = 0 has a discriminant of 0 and x = −1 twice. A negative value gives a pair of complex conjugate roots: x² + 2x + 5 = 0 has a discriminant of −16 and x = −1 ± 2i.
Substitute a root back into the equation to verify it: with x = 3 in 2x² − 4x − 6, the terms are 18 − 12 − 6 = 0. The roots also give the factored form a(x − r₁)(x − r₂), so 2x² − 4x − 6 = 2(x − 3)(x + 1). A negative leading coefficient only flips the parabola, as in −x² + 4 = 0 with roots −2 and 2. The value of a cannot be 0, because the equation would then be linear with at most one solution. Irrational roots are displayed rounded to eight decimal places.
| Input | Result |
|---|---|
1, -5, 6 | x = 3, 2 · Δ 1 |
2, -4, -6 | x = 3, -1 · Δ 64 |
1, 0, -2 | x = 1.41421356, -1.41421356 · Δ 8 |
1, 2, 1 | x = -1, -1 · Δ 0 |
1, 2, 5 | x = -1 ± 2i · Δ -16 |
-1, 0, 4 | x = -2, 2 · Δ 16 |
It decides the kind of answer before any square root is taken. x² − 2 = 0 has a discriminant of 8 and two real roots; x² + 2x + 1 = 0 has a discriminant of 0 and the repeated root −1; and x² + 2x + 5 = 0 has a discriminant of −16, so the roots are complex.
Substitute it back into the equation. With x = 3 in 2x² − 4x − 6, the terms are 18 − 12 − 6 = 0, so the root is correct. The roots also give the factored form a(x − r₁)(x − r₂), which means 2x² − 4x − 6 can be written as 2(x − 3)(x + 1).
Yes. A negative value of a simply flips the parabola: −x² + 4 = 0 has a discriminant of 16 and roots of −2 and 2. Only a = 0 is rejected, because a linear equation has at most one solution and the quadratic formula would divide by zero.
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