Quadratic Equation Solver
Solve ax²+bx+c=0 and find real or complex roots using the quadratic formula. Also shows vertex, axis of symmetry, and discriminant analysis.
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How It Works
Substitute a, b, c into x = (−b ± √(b²−4ac)) / 2a. Compute the discriminant first, then apply ± to find both roots.
x = (−b ± √(b²−4ac)) / 2ax²−5x+6=0: x=(5±√1)/2 → x=3 or x=2b²−4ac determines how many real solutions exist. Positive = 2 real roots; zero = 1 repeated; negative = 2 complex.
Δ = b²−4acΔ=25−24=1>0 → two real rootsThe vertex is at x=−b/2a and y = f(−b/2a). This is the minimum (opens up) or maximum (opens down) of the parabola.
Vertex: (−b/2a, f(−b/2a))x²−5x+6: vertex at (2.5, −0.25)Complex roots come in conjugate pairs: x = (−b ± i√(4ac−b²)) / 2a. The real part is −b/2a and imaginary part is √|Δ|/2a.
x = real ± imaginary·ix²+x+1=0: x=−0.5 ± 0.866iOnce roots r1 and r2 are found: ax²+bx+c = a(x−r1)(x−r2). Verify by expanding back.
f(x) = a(x−r1)(x−r2)x²−5x+6 = (x−2)(x−3)The y-intercept is always the constant c. Set x=0: f(0)=c. This is a quick check point for graphing.
f(0) = cx²−5x+6: y-intercept = 6Quick Reference
Common examples — verify instantly above.
Real roots
x²−5x+6=0
x=2, x=3
Real roots
2x²−x−3=0
x=1.5, x=−1
Double root
x²−4x+4=0
x=2 (double)
Complex
x²+1=0
x=±i
Complex
x²+x+1=0
complex roots
Vertex
x²−6x+5=0
vertex (3,−4)
Negative a
−x²+4=0
x=±2
Discriminant
x²+2x+2=0
Δ=−4 (complex)
Tips & Shortcuts
Always compute the discriminant first — it tells you the type of solution before doing any arithmetic.
If a is negative, the parabola opens downward and the vertex is a maximum point, not a minimum.
A quick factoring check: if b²−4ac is a perfect square, the equation factors nicely with integer or simple rational roots.
The sum of roots = −b/a and the product of roots = c/a. Use these to verify your solutions without expanding.
For complex roots, the real part −b/2a is the x-coordinate of the vertex. Complex roots always come in conjugate pairs.
If all coefficients share a common factor, divide them all first. 2x²−10x+12=0 simplifies to x²−5x+6=0.
Common Mistakes
Forgetting ± in the quadratic formula
The formula gives TWO solutions: one with + and one with −. Always compute both.
Computing √(b²) instead of √(b²−4ac)
The discriminant is b²−4ac, not just b². Subtract 4ac from b² before taking the square root.
Dividing only the numerator by 2a
Divide the entire (−b ± √Δ) by 2a, not just the √Δ part.
Thinking Δ < 0 means no solution exists
It means no REAL solution. Complex roots exist. x²+1=0 has solutions x=±i.
Setting a=0 and expecting a quadratic
If a=0, the equation is linear, not quadratic. The quadratic formula divides by 2a and fails when a=0.
Not simplifying √Δ before computing roots
Simplify the square root as much as possible first to get cleaner results and avoid arithmetic errors.
Frequently Asked Questions
x = (−b ± √(b²−4ac)) / 2a. It gives the roots of any quadratic equation ax²+bx+c=0.
If b²−4ac > 0: two real roots. If = 0: one repeated root. If < 0: two complex (imaginary) roots.
The vertex is the maximum or minimum point: x = −b/2a, y = f(−b/2a). It lies on the axis of symmetry.
Yes — when the discriminant < 0. The solutions are complex numbers involving i = √(−1).
When the discriminant = 0, both roots are equal: x = −b/2a. The parabola touches but does not cross the x-axis.
If roots are r1 and r2: ax²+bx+c = a(x−r1)(x−r2). Factoring works cleanly when roots are rational.
Completing the square rewrites ax²+bx+c as a(x−h)²+k where (h, k) is the vertex. This reveals the parabola's turning point directly: vertex at (−b/2a, c−b²/4a). The vertex form shows at a glance whether the parabola opens up (a>0, minimum at vertex) or down (a<0, maximum at vertex), and how far the axis of symmetry x=−b/2a is from the y-axis.
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