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Equation Solver Calculator — Linear & Quadratic Equation Solver

Solve linear and quadratic equations with step-by-step solutions

Format: ax + b = c (e.g., 2x + 3 = 7, 5x - 2 = 8)

Equation Solver: Linear and Quadratic Equation Calculator

Table of Contents - Equation Solver

How to Use This Calculator - Equation Solver

Select your equation type: Linear (equations like 2x + 3 = 7) or Quadratic (equations like x² + 2x - 3 = 0).

For linear equations, enter the equation in standard form with "x" as the variable. Example: "2x + 5 = 11" or "3x - 4 = 2x + 7".

For quadratic equations, enter in the form "ax² + bx + c = 0" or enter coefficients a, b, and c directly. Example: "x² + 5x + 6 = 0" or enter a=1, b=5, c=6.

Click "Solve" to see results. The output displays:

  • The solution(s) for x
  • Step-by-step breakdown of the solution process
  • For quadratics: discriminant value and nature of roots (real and distinct, real and equal, or complex)
  • Verification by substituting solution back into original equation

The Core Principle: Algebraic Manipulation

Solving equations means finding the value(s) of the variable that make the equation true. We do this by applying inverse operations to isolate the variable.

Linear equations (ax + b = c): One variable, one power (x¹). Always has exactly one solution. Solve by isolating x using inverse operations.

Quadratic equations (ax² + bx + c = 0): One variable, highest power is 2. Can have two solutions, one solution, or no real solutions (complex solutions). Solve by factoring, completing the square, or the quadratic formula.

The discriminant (b² - 4ac) determines the nature of quadratic solutions:

  • Positive: two distinct real roots
  • Zero: one repeated real root
  • Negative: two complex conjugate roots

Understanding these principles enables solving a vast range of practical problems that reduce to these equation forms.

How to Solve Equations Manually

Linear equation example: 3x + 7 = 22

Step 1: Subtract 7 from both sides 3x = 22 - 7 = 15

Step 2: Divide both sides by 3 x = 15 / 3 = 5

Verify: 3(5) + 7 = 15 + 7 = 22 ✓

Linear equation with variables on both sides: 5x - 3 = 2x + 9

Step 1: Subtract 2x from both sides 3x - 3 = 9

Step 2: Add 3 to both sides 3x = 12

Step 3: Divide by 3 x = 4

Quadratic equation by factoring: x² + 5x + 6 = 0

Step 1: Find factors of 6 that add to 5 2 × 3 = 6, and 2 + 3 = 5

Step 2: Factor (x + 2)(x + 3) = 0

Step 3: Set each factor to zero x + 2 = 0 → x = -2 x + 3 = 0 → x = -3

Quadratic equation using the quadratic formula: 2x² + 4x - 6 = 0

Formula: x = (-b ± √(b² - 4ac)) / (2a)

Where a = 2, b = 4, c = -6

Step 1: Calculate discriminant D = 4² - 4(2)(-6) = 16 + 48 = 64

Step 2: Apply formula x = (-4 ± √64) / (2 × 2) = (-4 ± 8) / 4

Step 3: Find both solutions x = (-4 + 8) / 4 = 1 x = (-4 - 8) / 4 = -3

Real-World Applications

Financial break-even analysis. Revenue = 50x (selling x units at $50). Costs = 1000 + 30x (fixed costs plus variable costs). Break-even: 50x = 1000 + 30x. Solve: x = 50 units.

Physics motion problems. An object thrown upward has height h = -16t² + 64t + 5. When does it hit the ground? Solve -16t² + 64t + 5 = 0.

Area and dimension problems. A rectangle has area 48 and length 2 more than width. If width = w: w(w + 2) = 48, or w² + 2w - 48 = 0. Solve to find w = 6.

Mixture problems. Mix x liters of 30% solution with 20 liters of 50% solution to get 40% mixture. Equation: 0.30x + 0.50(20) = 0.40(x + 20). Solve for x.

Time and rate problems. Two workers complete a job: one in 5 hours, one in 3 hours. Together: 1/5 + 1/3 = 1/t. Solve for t = 15/8 = 1.875 hours.

Scenarios People Actually Run Into

The no-solution case. 2x + 3 = 2x + 5. Subtracting 2x gives 3 = 5, which is false. This equation has no solution—the lines are parallel.

The infinite solution case. 2x + 4 = 2(x + 2). Expanding: 2x + 4 = 2x + 4. This is always true—any x works. The equation is an identity.

The negative discriminant. x² + 2x + 5 = 0. Discriminant = 4 - 20 = -16 < 0. No real solutions—the parabola doesn't cross the x-axis. Complex solutions exist: x = -1 ± 2i.

The hidden quadratic. An equation like √x = x - 2 isn't obviously quadratic. Squaring both sides: x = (x-2)² = x² - 4x + 4, or x² - 5x + 4 = 0. But check solutions—squaring can introduce extraneous roots.

The coefficient confusion. In 2x² - x = 3, you must first write in standard form: 2x² - x - 3 = 0. Now a = 2, b = -1 (not 1), c = -3.

Trade-Offs and Decisions People Underestimate

Factoring versus formula. Factoring is faster when it works but requires recognizing factor patterns. The quadratic formula always works but involves more calculation. Learn both.

Exact versus decimal answers. x = (-3 + √17)/2 is exact but hard to interpret. x ≈ 0.561 is easier to use but loses precision. Context determines which is better.

Checking solutions. Algebraic manipulations can introduce errors. Always substitute your answer back into the original equation to verify.

Simplification choices. Solving (x + 3)(x - 2) = 5 requires expanding first: x² + x - 6 = 5, so x² + x - 11 = 0. The original form isn't directly solvable.

When to use technology. Simple equations build understanding; complex ones justify calculators. Know when manual solving teaches you something and when it's just tedious.

Common Mistakes and How to Recover

Sign errors. The most common mistake. Moving terms across the equals sign changes their sign. 2x = 10 - 3x becomes 5x = 10 (adding 3x to both sides).

Forgetting both roots. √64 = ±8, not just 8. Quadratics typically have two solutions—don't stop at the first one.

Distributing incorrectly. 2(x + 3) = 2x + 6, not 2x + 3. Every term inside the parentheses gets multiplied.

Canceling incorrectly. (x + 2)/2 ≠ x + 1 (can't cancel the 2s). Correct simplification: (x + 2)/2 = x/2 + 1.

Squaring both sides without checking. If you square to eliminate a radical, you may introduce extraneous solutions. Always verify by substituting back into the original.

Related Topics

Systems of equations. Multiple equations with multiple variables. Solve using substitution or elimination. Example: 2x + y = 7 and x - y = 2.

Polynomial equations. Higher-degree equations (cubic, quartic, etc.). May have multiple solutions. Often require numerical methods for exact solutions.

Inequalities. Similar to equations but finding ranges. 2x + 3 > 7 gives x > 2. Remember: multiplying by negative reverses the inequality.

Absolute value equations. |x - 3| = 5 means x - 3 = 5 or x - 3 = -5, giving x = 8 or x = -2. Two cases to consider.

Exponential and logarithmic equations. Equations with x in exponents (2^x = 8) or in logarithms (log x = 2). Require different techniques.

How This Calculator Works

Linear equation solver:

  1. Parses the equation to identify left and right sides
  2. Extracts coefficient of x and constant term from each side
  3. Rearranges to ax = b form by moving terms
  4. Solves: x = b/a
  5. Generates step-by-step explanation

Quadratic equation solver:

  1. Parses equation or accepts coefficients a, b, c
  2. Ensures standard form ax² + bx + c = 0
  3. Calculates discriminant D = b² - 4ac
  4. Determines solution method based on D:
    • D > 0: two real solutions via quadratic formula
    • D = 0: one real solution (repeated root)
    • D < 0: two complex solutions
  5. Applies quadratic formula: x = (-b ± √D) / (2a)
  6. Simplifies radicals where possible
  7. Generates step-by-step explanation with verification

All calculations happen locally in your browser.

FAQs

What's the difference between an equation and an expression?

An equation has an equals sign and can be solved for unknown values. An expression (like 2x + 3) has no equals sign—it can be simplified but not "solved."

Why does a quadratic have two solutions?

A quadratic equation represents a parabola. The solutions are where it crosses the x-axis. Since a parabola can cross at two points, one point, or not at all, quadratics can have two, one, or no real solutions.

What does a negative discriminant mean?

The discriminant (b² - 4ac) being negative means the parabola doesn't cross the x-axis—no real solutions exist. However, complex solutions (involving i = √-1) do exist.

When should I use the quadratic formula versus factoring?

Try factoring first for simple equations with integer coefficients. If factors aren't obvious within 30 seconds, use the quadratic formula—it always works.

How do I verify my solution?

Substitute your answer back into the original equation. Both sides should equal the same value. If they don't, you made an error somewhere.

What if my equation has no solution?

Some equations (like x + 1 = x + 2) have no solution—they're contradictions. This is a valid result, not an error.

Can this calculator solve systems of equations?

This calculator handles single equations. Systems of equations (multiple equations with multiple unknowns) require different methods like substitution or elimination.

What about equations with fractions?

Multiply both sides by the LCD (lowest common denominator) to clear fractions, then solve the resulting equation. Example: x/2 + x/3 = 5 becomes 3x + 2x = 30, so x = 6.

How do I solve equations with absolute values?

An equation like |x - 3| = 5 means either x - 3 = 5 or x - 3 = -5. Solve both cases: x = 8 or x = -2. Check both solutions in the original equation.

What if my equation has radicals?

Isolate the radical, then square both sides to eliminate it. But squaring can introduce extraneous solutions, so always verify answers in the original equation. Example: √x = x - 2 becomes x = (x-2)², which simplifies to x² - 5x + 4 = 0.

How do I solve systems of two equations?

Use substitution (solve one equation for a variable, substitute into the other) or elimination (add/subtract equations to eliminate a variable). The solution is where both equations are satisfied simultaneously.

When should I use the quadratic formula versus completing the square?

The quadratic formula always works and is usually faster. Completing the square is useful when you need vertex form, when deriving the formula, or when the leading coefficient is 1 and the linear coefficient is even.