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Quadratic Equation Solver with Graph and Steps — A Gentle, Complete Guide

Imagine you’re solving a problem in algebra class or modeling a real-world curve — the quadratic equation keeps popping up:
ax² + bx + c = 0. You plug in the coefficients and want the answers fast, but you also want to understand them. This guide walks you through everything: what the solver does, why it matters, the formulas, exact step-by-step solving methods, how the graph ties in, common errors, and the handful of facts that make you look confident in class.

I’ve taught and worked with quadratics for decades — here’s a clear, no-nonsense explanation you can use immediately.

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What is this?

A quadratic equation is any equation that can be written as:

a x² + b x + c = 0

where a, b, and c are numbers and a ≠ 0.
A Quadratic Equation Solver takes a, b, c and returns:

• the discriminant D,
  
• the two roots (solutions) x₁ and x₂ (real or complex),
  
• the vertex and axis of symmetry, and
  
• a graph of the parabola y = a x² + b x + c showing roots and vertex.

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The Core Formulae

Discriminant:

D = b² − 4ac

Quadratic formula (roots):

x = (-b ± √D) / (2a)

Vertex (point of minimum/maximum):

x_vertex = -b / (2a)

y_vertex = a*(x_vertex)² + b*(x_vertex) + c

Axis of symmetry: x = -b/(2a)

Sum and Product of roots (Vieta):

x₁ + x₂ = -b / a

x₁ * x₂ = c / a

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How the solver works

1. Input validation
   
   • Check a, b, c are numbers.
     
   • If a = 0, reduce to a linear equation b x + c = 0 (root x = −c/b if b ≠ 0); otherwise it’s not quadratic.
     
2. Compute discriminant
   
   • D = b² − 4ac.
     
3. Decide nature of roots
   
   • If D > 0: two distinct real roots.
     
   • If D = 0: one repeated real root (double root).
     
   • If D < 0: two complex conjugate roots.
     
4. Compute roots carefully
   

If D ≥ 0: √D is real; compute

x₁ = (-b + √D) / (2a)

x₂ = (-b − √D) / (2a)

If D < 0: set √D = i√|D| and compute complex roots:

x₁ = (-b / (2a)) + i (√|D| / (2a))

x₂ = (-b / (2a)) − i (√|D| / (2a))

For numerical stability (when b is large), use the alternative stable method:

q = -0.5 * (b + sign(b) * √D)

x₁ = q / a

x₂ = c / q

•  This avoids catastrophic cancellation when b and √D nearly cancel.
  

5. Compute vertex and y-intercept
   
   • x_vertex = -b / (2a)
     
   • y_vertex = f(x_vertex)
     
   • y_intercept = c (point (0, c))
     
6. Prepare graph
   
   • Plot y = a x² + b x + c.
     
   • Mark roots and vertex, label axis and intercepts.
     
   • Choose x-range to comfortably include vertex and roots (for example, center on x_vertex with ± some span).
     
7. Output
   
   • Discriminant, root1, root2, vertex coordinates, axis of symmetry, and a clear parabola graph.
     

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Manual solving — exact step-by-step examples

Example A — Two distinct real roots

Solve x² − 3x + 2 = 0 (a=1, b=−3, c=2)

1. Compute discriminant:
   
   • b² = (−3)² = 9
     
   • 4ac = 4×1×2 = 8
     
   • D = 9 − 8 = 1
     
2. √D = 1
   
3. Roots:
   
   • x₁ = (−b + √D)/(2a) = (−(−3) + 1)/(2×1) = (3 + 1)/2 = 4/2 = 2
     
   • x₂ = (−b − √D)/(2a) = (3 − 1)/2 = 2/2 = 1
     
4. Vertex:
   
   • x_vertex = −b/(2a) = 3/2 = 1.5
     
   • y_vertex = f(1.5) = (1.5)² − 3(1.5) + 2 = 2.25 − 4.5 + 2 = −0.25
     

Graph crosses the x-axis at x = 1 and x = 2, vertex at (1.5, −0.25).

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Example B — Repeated root

Solve x² − 4x + 4 = 0 (a=1, b=−4, c=4)

1. b² = 16, 4ac = 16, so D = 0.
   
2. Single root:
   
   • x = −b/(2a) = 4/2 = 2
     
   • It’s a double root; graph just touches x-axis at x=2 (vertex on x-axis).
     

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Example C — Complex roots

Solve x² + 1 = 0 (a=1, b=0, c=1)

1. b² = 0, 4ac = 4, so D = 0 − 4 = −4.
   
2. |D| = 4, √|D| = 2.
   

Roots:

x = (−0 ± i*2)/(2*1) = ± i

• So x₁ = i, x₂ = −i.
  

Graph has no x-intercepts; vertex at (0, 1) and opens upward.