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Equation Solver

Solve linear equations (ax+b=0), quadratic equations (ax²+bx+c=0 with discriminant Δ and complex roots), and 2x2 systems with step-by-step solutions.

Roots & Solution
x₁ = 3
x₂ = 2
Step-by-Step Solution:
Equation: 1x² + (-5)x + (6) = 0
Compute discriminant Δ = b² - 4ac = (-5)² - 4(1)(6) = 1
Δ > 0: Two distinct real roots.
x₁ = (-(-5) + √1) / (2 · 1) = 3
x₂ = (-(-5) - √1) / (2 · 1) = 2
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What Is a Equation Solver?

The Equation Solver provides step-by-step solutions for fundamental algebraic equations: linear equations (ax + b = 0), quadratic equations (ax² + bx + c = 0) including real and complex conjugate roots, and 2×2 systems of linear equations using Cramer’s Rule.

How to Use This Calculator

  1. Select the equation type: Quadratic (ax² + bx + c = 0), Linear (ax + b = 0), or 2×2 System.
  2. Enter the numeric coefficients (a, b, c).
  3. Review the exact real or complex roots and the complete step-by-step algebraic deduction.

Quadratic Formula & Discriminant

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \Delta = b^2 - 4ac

If Δ > 0, there are two distinct real roots; if Δ = 0, one repeated root; if Δ < 0, two complex conjugate roots p ± qi.

Worked Example

Scenario: Solving x² - 5x + 6 = 0

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

Discriminant: Δ = (-5)² - 4(1)(6) = 25 - 24 = 1.

Apply formula: x = (5 ± √1) / 2 = (5 ± 1) / 2.

Roots: x₁ = 3, x₂ = 2.

Result: x₁ = 3, x₂ = 2.

Tips & Key Notes

  • If the discriminant Δ is positive, the quadratic graph crosses the x-axis at two distinct points.
  • If Δ is zero, the vertex touches the x-axis at exactly one point.
  • If Δ is negative, the graph never crosses the x-axis, producing complex conjugate roots.

Frequently Asked Questions

What happens when the discriminant (Δ) is negative?

When Δ < 0, the equation has no real solutions. Instead, it has two complex conjugate roots containing the imaginary unit i (where i = √(-1)).

How does Cramer’s Rule solve a 2x2 system of equations?

Cramer’s Rule computes the determinant of the coefficient matrix D = a₁b₂ - a₂b₁. If D ≠ 0, the unique solutions are x = D_x / D and y = D_y / D.

What if the leading coefficient a is 0 in a quadratic equation?

If a = 0, the equation reduces to a first-degree linear equation (bx + c = 0), which has a single solution x = -c/b.

Can this solver handle decimal and negative coefficients?

Yes, all positive, negative, integer, and decimal coefficients are supported.

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