Equation Solver
The tool below solves an equation in one variable. Enter the equation and the calculator walks through the work step by step — isolating the variable, combining like terms, and applying the right technique for the equation type.
Type naturally or click ⌨ in the box to use the math keyboard.
Examples: 2x + 5 = 13 for a linear equation, x^2 - 4 = 0 for a quadratic.
Use any single letter for the variable.
To solve an equation is to find every value of the variable that makes the equation true. For \(2x + 5 = 13\), the only solution is \(x = 4\), because substituting back gives \(2(4) + 5 = 13\). The right strategy depends on the equation type — linear, quadratic, or with fractions.
Worked Examples
Three examples worked by hand — the same logic the calculator uses, just spelled out so you can follow it on paper.
Example 1: Linear equation — solve \(3x - 7 = 8\)
Use inverse operations to isolate \(x\). Add 7 to both sides to undo the subtraction:
\[3x = 15\]
Divide both sides by 3 to undo the multiplication:
\[x = 5\]
Check: \(3(5) - 7 = 15 - 7 = 8\). ✓
Example 2: Equation with a fraction — solve \(\dfrac{x}{4} + 2 = 5\)
First, subtract 2 from both sides:
\[\dfrac{x}{4} = 3\]
Then multiply both sides by 4 to undo the division:
\[x = 12\]
Alternatively, you could multiply both sides by 4 at the start to clear the fraction, then solve normally.
Example 3: Quadratic by factoring — solve \(x^2 - 5x + 6 = 0\)
Look for two numbers that multiply to 6 and add to \(-5\). The pair \((-2, -3)\) works:
\[x^2 - 5x + 6 = (x - 2)(x - 3) = 0\]
By the zero product property, one of the factors must be zero:
\[x - 2 = 0 \quad \text{or} \quad x - 3 = 0\] \[x = 2 \quad \text{or} \quad x = 3\]
Quadratics typically have two solutions. If the quadratic doesn't factor cleanly, fall back to the quadratic formula.
What Does it Mean to Solve an Equation?
To solve an equation is to find every value of the variable that makes the equation true. For \(2x + 5 = 13\), the only value of \(x\) that works is \(x = 4\), because substituting 4 back in gives \(2(4) + 5 = 13\). ✓
The strategy depends on what kind of equation you have:
- Linear equation (like \(2x + 5 = 13\)): isolate the variable using inverse operations.
- Quadratic equation (like \(x^2 - 4 = 0\)): factor when you can, otherwise use the quadratic formula.
- Equations with fractions: clear denominators by multiplying through.
- Equations with parentheses: distribute first, then collect like terms.
The calculator above identifies the equation type and shows each step so you can follow the logic, not just copy the answer.
Try These Examples
2x + 5 = 13— one-step linear3y - 7 = 20— different variablet/4 + 3 = 7— equation with a fraction5(n - 2) = 15— needs distributionz^2 - 4 = 0— quadratic by factoringx^2 + 5x + 6 = 0— factorable quadratic
Inverse Operations (the Toolkit)
A quick reference for isolating a variable. Whatever you do to one side, do to the other:
| To undo | Do this to both sides |
|---|---|
| \(+a\) | subtract \(a\) |
| \(-a\) | add \(a\) |
| \(\times a\) | divide by \(a\) |
| \(\div a\) | multiply by \(a\) |
| squaring | take the square root (both \(\pm\)) |
| square root | square both sides |
For quadratics that don't factor cleanly, the quadratic formula gives the solutions to \(ax^2 + bx + c = 0\):
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Tips for Using the Calculator
- Use one variable per equation — any single letter works (\(x\), \(y\), \(t\), \(n\), …)
- For powers, use
^:x^2means \(x^2\) - For fractions, use
\frac{a}{b}or justa/b - For square roots, use
\sqrt{x} - The equation must include an
=sign — otherwise it's an expression, not an equation
If you want to factor a polynomial without solving an equation, use the Factoring Calculator. For systems of equations with more than one variable, see Systems of Linear Equations.
Frequently Asked Questions
Why does my quadratic have two solutions?
A quadratic equation \(ax^2 + bx + c = 0\) has a degree-2 polynomial on one side, and by the Fundamental Theorem of Algebra a degree-\(n\) polynomial has up to \(n\) solutions. Visually, the graph of a parabola can cross the x-axis at up to two points — each crossing is a real solution.
When do I use the quadratic formula vs. factoring?
Try factoring first — it's usually faster when the quadratic factors over the integers. If you can't find a factoring pair quickly, or the leading coefficient is messy, switch to the quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It always works, even when factoring doesn't.
What does it mean if my equation has no solution?
Some equations are contradictions — they're false no matter what value you plug in. For example, \(x + 1 = x\) simplifies to \(1 = 0\), which is never true. There's no solution. For quadratics, "no real solution" means the discriminant \(b^2 - 4ac\) is negative; there are still two complex solutions, but none on the real number line.
What's an "extraneous" solution?
An extraneous solution is a value that satisfies a modified version of your equation but not the original. They often appear after squaring both sides or multiplying through by a variable expression. Always check your answers in the original equation — if a value doesn't work there, discard it.
How do I check my answer?
Substitute the solution back into the original equation. Compute both sides separately. If both sides are equal, your solution is correct. This catches arithmetic errors and also extraneous solutions. The calculator above doesn't skip the check — neither should you.