🔢 Math Calculator
Equation Solver
Solve linear equations (ax + b = c), quadratic equations (ax² + bx + c = 0), and systems of two equations instantly with full step-by-step solutions.
Solve: ax + b = c → x = (c − b) / a
Enter values above to see the equation
Solution: x =
Enter values above
Solve: ax² + bx + c = 0 using the Quadratic Formula
Enter values above to see the equation
Solutions
Enter values above
Solve simultaneously: a₁x + b₁y = c₁ and a₂x + b₂y = c₂
Equation 1: a₁x + b₁y = c₁
Equation 2: a₂x + b₂y = c₂
Solutions
Enter values above
Solve: a/b = c/x → x = (b × c) / a
Enter values above
Solution: x =
Enter values above
📖 How to Use the Equation Solver
1
Choose a tab — Linear, Quadratic, System of 2 Equations, or Proportion.
2
Linear equations — enter a, b, c for the equation ax + b = c and get x instantly.
3
Quadratic equations — enter a, b, c for ax² + bx + c = 0 and get both roots x₁ and x₂.
4
System of equations — enter coefficients for two equations and solve for x and y simultaneously.
5
Tap Steps to see the full working, or Copy to copy the answer.
⚡ Why Use Our Equation Solver?
=
4 Equation Types
Linear, quadratic, systems, and proportions — all covered.
📋
Step-by-Step
Full working shown — perfect for students and homework.
⚡
Real-Time
Solutions update instantly as you type coefficients.
📱
Mobile-Friendly
Works perfectly on all phones, tablets, and desktops.
🔢
Complex Roots
Handles quadratics with no real roots — shows complex solutions.
🆓
100% Free
No sign-up — completely free forever.
❓ Frequently Asked Questions
For ax + b = c: isolate x by subtracting b from both sides, then divide by a. Result: x = (c − b) ÷ a. For example, 2x + 3 = 11 → x = (11−3)/2 = 4.
For ax² + bx + c = 0, the solutions are: x = (−b ± √(b²−4ac)) / (2a). The expression b²−4ac is called the discriminant. If it is positive, there are two real roots. If zero, one root. If negative, the roots are complex numbers.
We use Cramer's Rule (determinant method). For a₁x+b₁y=c₁ and a₂x+b₂y=c₂: calculate determinant D = a₁b₂ − a₂b₁. Then x = (c₁b₂−c₂b₁)/D and y = (a₁c₂−a₂c₁)/D. If D=0, the system has no unique solution.