Equation Solver — Solve Linear, Quadratic, and Polynomial Equations

Types of Equations the Solver Handles

How to Use the Equation Solver

1. Open the equation solver.
2. Enter your equation in the input field using standard notation. Use ^ for powers (x^2 = x²), * for multiplication, and / for division.
3. Examples of valid input: 3x + 5 = 14, x^2 - 5x + 6 = 0, 2x^3 - x^2 + 3x - 2 = 0.
4. Click Solve. The solver returns all roots and shows the step-by-step method used.
5. Check whether roots are real or complex, exact (fractions/surds) or decimal approximations.

Solving Linear Equations

A linear equation (degree 1) has exactly one solution. The goal is to isolate the variable using inverse operations — always performing the same operation to both sides to maintain equality.

Example: Solve 5x − 3 = 2x + 9

1. Subtract 2x from both sides: 3x − 3 = 9
2. Add 3 to both sides: 3x = 12
3. Divide both sides by 3: x = 4

Verify: 5(4) − 3 = 17 and 2(4) + 9 = 17 ✓

Always verify by substituting the solution back into the original equation. This catches arithmetic errors and confirms the solution is correct.

Solving Quadratic Equations

Method 1: Factorisation

For x² − 5x + 6 = 0, find two numbers that multiply to +6 and add to −5: those are −2 and −3.

Factor: (x − 2)(x − 3) = 0

By the zero product property: x − 2 = 0 → x = 2, or x − 3 = 0 → x = 3.

Solutions: x = 2, x = 3.

Method 2: The Quadratic Formula

For ax² + bx + c = 0, the quadratic formula always gives the solutions:

x = (−b ± √(b² − 4ac)) / 2a

For x² − 5x + 6 = 0 (a=1, b=−5, c=6): x = (5 ± √(25−24)) / 2 = (5 ± 1) / 2, giving x = 3 or x = 2. ✓

The Discriminant

Comparison of Solution Methods

Complex Roots and What They Mean

When the discriminant is negative, the quadratic has no real roots. The roots are complex numbers of the form a ± bi, where i = √(−1). For example, x² + 4 = 0 has roots x = ±2i.

Complex roots always come in conjugate pairs (a + bi and a − bi) for polynomials with real coefficients. In most school-level algebra, complex roots indicate that the equation has no solution in the real numbers — but in engineering, physics, and advanced mathematics, complex roots are physically meaningful (AC circuit analysis, quantum mechanics, control theory).

The Fundamental Theorem of Algebra

Every polynomial of degree n has exactly n roots in the complex numbers (counting multiplicity). A quadratic has 2 roots; a cubic has 3; a quartic has 4. Some may be complex; some may be repeated. This theorem, first proved by Gauss in 1799, guarantees that the equation solver will always find all roots — either real or complex.

For real-coefficient polynomials, complex roots always occur in conjugate pairs. This means a cubic polynomial must have at least one real root (since the third complex root cannot be paired). Similarly, a quartic may have 0, 2, or 4 real roots.

Solving Simultaneous Equations

Substitution method

Rearrange one equation to express one variable in terms of the other, then substitute into the second equation.

Example: x + y = 5 and 2x − y = 1.

1. From equation 1: y = 5 − x.
2. Substitute into equation 2: 2x − (5 − x) = 1 → 3x − 5 = 1 → x = 2.
3. Back-substitute: y = 5 − 2 = 3.
4. Solution: x = 2, y = 3.

Elimination method

Add or subtract the equations to eliminate one variable.

From the same example, adding the two equations: (x + y) + (2x − y) = 5 + 1 → 3x = 6 → x = 2. Then y = 3 as before.

Practical Applications

Break-even analysis (business)

Revenue and cost functions can be set equal to find the break-even point: Revenue(x) = Cost(x). For R(x) = 50x and C(x) = 30x + 400, solve 50x = 30x + 400 → 20x = 400 → x = 20 units to break even.

Projectile height (physics)

A ball is thrown upward with initial velocity 20 m/s from a height of 5 m. Its height is h = −4.9t² + 20t + 5. To find when it hits the ground: −4.9t² + 20t + 5 = 0. Use the quadratic formula with a = −4.9, b = 20, c = 5.

Geometry problems

Finding the dimensions of a rectangle with area 24 m² and perimeter 20 m: let width = w, length = l. Then l × w = 24 and 2(l + w) = 20 → l + w = 10 → l = 10 − w. Substituting: w(10 − w) = 24 → w² − 10w + 24 = 0 → (w − 4)(w − 6) = 0, giving dimensions 4 m × 6 m.

Common Mistakes

Not moving everything to one side before using the quadratic formula

The quadratic formula requires the equation to be in the form ax² + bx + c = 0. If you have x² − 3x = 10, first rearrange to x² − 3x − 10 = 0, then identify a=1, b=−3, c=−10.

Sign errors in the discriminant

−4ac is subtracted from b², not b. If b = −3, then b² = 9 (positive). If c = −10 and a = 1, then −4ac = −4(1)(−10) = +40. Many errors come from treating a negative c as making −4ac negative.

Forgetting the ± in the quadratic formula

The quadratic formula gives TWO solutions: one for + and one for −. Both are valid solutions of the equation. Forgetting the ± produces only one root where there should be two.

Common Questions

Can the solver handle equations with no solution?

Yes. If a linear equation reduces to a false statement (e.g., 0 = 5), the solver reports "no solution." If it reduces to a true statement (e.g., 0 = 0), the solver reports "infinitely many solutions." For quadratics with negative discriminant, the solver reports complex roots.

What is the difference between a root and a solution?

These terms are used interchangeably in most contexts. "Root" typically refers to the value(s) of x that make a polynomial equal to zero; "solution" refers to the value(s) that satisfy an equation. For x² − 5x + 6 = 0, the roots of the polynomial x² − 5x + 6 are x = 2 and x = 3, which are also the solutions of the equation.

Why does the solver show the answer as a decimal and as an exact form?

For irrational roots (those involving surds like √5), the exact form (e.g., (3 + √5)/2) is mathematically precise but not easy to use numerically. The decimal approximation (e.g., 2.618...) is easier to use in further calculation but loses some precision. Both are shown so you can choose the appropriate form for your use.