What This Quadratic Equation Solver Does

Feature	Description
Solve ax² + bx + c = 0	Standard quadratic equation
Real Roots	Two distinct or one double root
Complex Roots	Handles negative discriminants
Discriminant Analysis	Shows Δ = b² – 4ac
Vertex Calculation	Finds parabola vertex point
Step-by-Step	Shows complete solution process
Quick Examples	One-click sample equations
Linear Equations	Handles cases where a = 0

The Quadratic Formula

x = [-b ± √(b² – 4ac)] / (2a)

Discriminant Cases

Discriminant (Δ)	Root Type	Example
Δ > 0	Two distinct real roots	x² – 3x + 2 = 0 → x = 1, 2
Δ = 0	One real root (double)	x² – 4x + 4 = 0 → x = 2
Δ < 0	Two complex conjugate roots	x² + 2x + 5 = 0 → x = -1 ± 2i

Examples Included

Equation	a	b	c	Roots
x² – 3x + 2 = 0	1	-3	2	x = 1, 2
x² + 2x – 3 = 0	1	2	-3	x = 1, -3
2x² + 5x – 3 = 0	2	5	-3	x = 0.5, -3
x² – 4x + 4 = 0	1	-4	4	x = 2 (double)
x² – 9 = 0	1	0	-9	x = 3, -3
x² + 4 = 0	1	0	4	x = ±2i
x² + 2x + 5 = 0	1	2	5	x = -1 ± 2i
3x² + 6x + 3 = 0	3	6	3	x = -1 (double)

Step-by-Step Example

Solve: x² – 5x + 6 = 0

text

Step 1: Identify coefficients
a = 1, b = -5, c = 6

Step 2: Calculate discriminant
Δ = b² - 4ac
Δ = (-5)² - 4(1)(6)
Δ = 25 - 24 = 1

Step 3: Apply quadratic formula
x = [-(-5) ± √1] / (2×1)
x = [5 ± 1] / 2
x₁ = (5 - 1)/2 = 2
x₂ = (5 + 1)/2 = 3

Solutions: x = 2, 3

Vertex Information

For quadratic equation ax² + bx + c:

• Vertex x-coordinate: x = -b/(2a)
• Vertex y-coordinate: f(x) = a(x)² + b(x) + c

Special Cases Handled

Case	Description	Solution
a = 0, b ≠ 0	Linear equation	x = -c/b
a = 0, b = 0, c = 0	Identity	All real numbers
a = 0, b = 0, c ≠ 0	Contradiction	No solution

Keyboard Shortcuts

• Ctrl/Cmd + Enter — Solve equation
• Ctrl/Cmd + R — Reset to default