Core answer: To solve ax²+bx+c=0 (a≠0): ① compute the discriminant Δ=b²−4ac; ② apply the quadratic formula x=(−b±√Δ)/2a. Δ>0 → two distinct real roots, Δ=0 → one repeated root, Δ<0 → a conjugate complex pair. Example: 2x²−7x+3=0, Δ=49−24=25, x=(7±5)/4 → x₁=3, x₂=0.5.
The quadratic formula and discriminant
For the general form ax² + bx + c = 0 (a ≠ 0):
Δ = b² − 4ac, x = (−b ± √Δ) ÷ 2a
The discriminant tells you how many real roots exist before you even take the square root — always compute it first.
Three cases of the discriminant
| Discriminant | Roots | Parabola vs x-axis | Example |
|---|---|---|---|
| Δ > 0 | two distinct real roots | 2 intersections | x²−3x+2=0 → 1, 2 |
| Δ = 0 | one repeated root | 1 (tangent) | x²−4x+4=0 → x=2 |
| Δ < 0 | no real roots; complex pair | 0 | x²+2x+5=0 → −1±2i |
Derivation by completing the square
The formula comes from completing the square, not from thin air:
- Divide by a: x² + (b/a)x + c/a = 0
- Move the constant: x² + (b/a)x = −c/a
- Add (b/2a)² to both sides: (x + b/2a)² = (b²−4ac)/4a²
- Take the square root and rearrange → x = (−b ± √Δ)/2a
The same technique rewrites y=ax²+bx+c in vertex form y=a(x−h)²+k with vertex (−b/2a, c−b²/4a).
Worked example 1: two distinct roots
Solve 2x² − 7x + 3 = 0:
- a=2, b=−7, c=3
- Δ = (−7)² − 4×2×3 = 49 − 24 = 25 > 0
- x = (7 ± 5) ÷ 4
- x₁ = 12/4 = 3, x₂ = 2/4 = 0.5
Check: 2×9 − 21 + 3 = 0 ✓. This one also factors as (2x−1)(x−3)=0 — faster when you spot it.
Worked example 2: repeated and complex roots
Repeated root: x² − 4x + 4 = 0, Δ = 16 − 16 = 0, x = 4/2 = 2 (double root), i.e. (x−2)²=0.
Complex roots: x² + 2x + 5 = 0, Δ = 4 − 20 = −16 < 0 — no real solution; in complex numbers x = (−2 ± 4i)/2 = −1 ± 2i.
Checking roots with Vieta's formulas
You can verify roots without re-solving:
| Relation | Formula | Check (x₁=3, x₂=0.5) |
|---|---|---|
| Sum of roots | x₁+x₂ = −b/a | 3.5 = 7/2 ✓ |
| Product of roots | x₁·x₂ = c/a | 1.5 = 3/2 ✓ |
Reverse uses: find the other root given one, build an equation from two desired roots, or evaluate symmetric expressions like x₁²+x₂² = (x₁+x₂)² − 2x₁x₂ without solving.
Common mistakes and myths
- Copying the sign of b wrong: in 2x²−7x+3=0, b=−7 so the numerator has −b=+7. Write down a, b, c with their signs before substituting.
- "Δ<0 means I made an error" — no, it simply means the parabola never crosses the x-axis; a conjugate complex pair still exists.
- Applying the formula when a=0 — then it is a linear equation; the denominator 2a would be zero. Just solve by rearranging.
- Forgetting the ± — when Δ>0 there are always two roots; dropping one loses half the answer.
Use the [Quadratic Equation Calculator](/c/math/quadratic) for instant roots, discriminant and vertex, with the parabola graphed alongside.