Help With Algebra Problems

What is the value of x in your equation?

Enter a linear or quadratic equation and get the solution instantly, along with each step of the working. Covers one-variable equations, systems of two equations, and quadratic expressions.

Updated July 2026 · How this works

Example calculation — edit any field to use your own numbers

Worth knowing
How It Works
The formula, explained simply

Think of a linear equation as a set of scales. On one side you have an unknown weight labelled x, multiplied by some factor and with a fixed weight added or removed. On the other side sits a known total. Solving the equation means figuring out what x weighs so both sides balance — and the process is just moving weights around systematically until x stands alone.

Quadratic equations are different in kind. Instead of a straight line, the relationship between x and the result traces a U-shaped curve called a parabola. Asking for the solution means asking where that curve crosses the horizontal axis — and because a curve can cross a straight line at zero, one, or two points, quadratics can have zero, one, or two real answers. The quadratic formula captures all three outcomes in one expression by using the discriminant to determine which case applies before committing to arithmetic.

The verification step in this tool is not decorative. It substitutes your answer back into the original equation and checks the result against zero or c within a small tolerance. Floating-point arithmetic means the check may return 0.0000001 rather than exactly 0 — that is normal rounding behaviour, not an error. Any residual smaller than 0.0001 is treated as verified.

When To Use This
Right tool, right situation

Use this tool when you have a specific equation and need to verify your hand-computed answer, when you are working through a multi-step problem and algebra is just one stage, or when you are double-checking a coefficient setup before writing up a solution. It is useful any time you need to be certain of the mechanics without spending time on arithmetic.

It is also useful outside pure maths. Engineers setting up force balance equations, students modelling projectile motion, and anyone building a cost model that involves a squared term can enter their rearranged equation directly and get roots in seconds. The step-by-step output shows the exact arithmetic chain, making it straightforward to trace back to the original physical or financial setup.

Where this tool is not appropriate: simultaneous equations with more than two unknowns, polynomial equations of degree three or higher (cubics, quartics), equations involving trigonometric or logarithmic terms, or inequalities where the answer is a range rather than a point. For those, a computer algebra system or specialist solver is the right choice. Trying to force a cubic into a quadratic form by ignoring terms produces roots that are mathematically wrong and will fail verification.

Common Mistakes
Why results sometimes look wrong

Mistake 1: Putting c on the wrong side for quadratics. The quadratic formula requires the equation in the form ax² + bx + c = 0 — everything on the left, zero on the right. If your equation reads x² + 3x = 10, you must rearrange to x² + 3x - 10 = 0 before entering coefficients. Entering c = 10 instead of c = -10 produces wrong roots and a verification failure.

Mistake 2: Treating the sign of b as separate from the coefficient. In the equation x² - 7x + 12 = 0, the coefficient b is -7, not 7. Entering 7 instead of -7 changes the discriminant from 25 to -23, flipping the problem from two real roots to no real solution. The quadratic formula applies b² internally, so the sign affects only the ±b term but gets the answer completely wrong.

Mistake 3: Expecting exact integers when the answer is irrational. Not all quadratic roots are whole numbers or simple fractions. The equation x² + x - 1 = 0 has roots at (-1 + √5) / 2 ≈ 0.618034 and (-1 - √5) / 2 ≈ -1.618034. These are exact algebraic numbers displayed as decimals. The verification step will confirm they satisfy the equation even though they look messy — that is the expected result, not a calculation error.

The Math
Worked examples and deeper derivation

For a linear equation ax + b = c, the single operation is x = (c - b) / a. This is defined everywhere a is not zero. When a equals zero and b does not equal c, the equation is a contradiction (0 = non-zero) with no solution. When a equals zero and b equals c, every real number satisfies it — infinitely many solutions. This tool catches the a = 0 case and reports it directly.

For a quadratic ax² + bx + c = 0, the quadratic formula is x = (-b ± √(b² - 4ac)) / 2a. The discriminant D = b² - 4ac classifies the result: D greater than 0 yields two distinct real roots, D equal to 0 yields one repeated root at x = -b / 2a, and D less than 0 yields two complex conjugate roots with no real part that lies on the number line. The tool halts at D less than 0 and explains why.

Floating-point precision matters when coefficients are large or when the discriminant is very close to zero. A discriminant computed as -0.000000001 due to rounding may represent a genuine double root, not a complex pair. For classroom and everyday use the precision here is more than sufficient, but for numerical analysis at scale, interval arithmetic or symbolic computation tools handle the edge cases more rigorously.

Checking homework: a linear equation
Equation type: Linear, a = 3, b = 5, c = 11
The solver finds x = 2, meaning 3(2) + 5 = 11 checks out perfectly. This is the fastest way to confirm your answer before submitting — the verification step shows the substitution so you can follow the reasoning, not just trust a number.
Factoring a quadratic with two real roots
Equation type: Quadratic, a = 1, b = -5, c = 6
The quadratic formula gives x₁ = 3 and x₂ = 2. Both values satisfy the equation x² - 5x + 6 = 0. If you were trying to factor this by hand as (x - 3)(x - 2), this confirms those factors are correct.
A physics student solving a projectile equation
Equation type: Quadratic, a = -4.9, b = 20, c = -15
The two roots represent the two times (in seconds) when a projectile is at a height of 15 metres given an initial velocity of 20 m/s. This tool handles non-integer and negative leading coefficients cleanly — the verification step shows the substitution matches zero within tolerance, confirming the physics is right.
Expert Unlock
The thing most explanations skip

The quadratic formula is numerically unstable when one root is much larger in magnitude than the other — specifically when |b| is large relative to √(b² - 4ac). In that case, computing (-b + √D) can lose significant digits through catastrophic cancellation. The stable fix is to compute the larger root first using the formula with the sign that avoids cancellation, then recover the smaller root via Vieta’s formula: x₁ × x₂ = c/a. For textbook problems with modest coefficients this never matters. For scientific computing with coefficients spanning several orders of magnitude, it can change the last several digits of the smaller root.

What does your algebra result actually mean?

How do I solve ax + b = c step by step?
To solve ax + b = c, subtract b from both sides to get ax = c - b, then divide both sides by a to isolate x = (c - b) / a. This works as long as a is not zero — if a equals zero there is no x term and the equation either has no solution or infinitely many, depending on whether b equals c.
What is the quadratic formula and when do I use it?
The quadratic formula solves any equation in the form ax² + bx + c = 0 by computing x = (-b ± √(b² - 4ac)) / 2a. Use it when factoring is not obvious or when coefficients are decimals. The term inside the square root, called the discriminant (b² - 4ac), tells you how many real solutions exist: positive means two solutions, zero means one repeated solution, negative means no real solutions.
Why does my quadratic equation have no real solution?
A quadratic equation has no real solutions when the discriminant b² - 4ac is negative, because you cannot take the square root of a negative number in the real number system. The equation still has two solutions, but they are complex numbers involving the imaginary unit i. This happens with equations whose parabola never crosses the x-axis, such as x² + 1 = 0.

Need something this doesn't cover?

Suggest a tool — we'll build it →