Quadratic Equation Solver

Type in a, b, and c and you get the two roots, the discriminant, the vertex, and the quadratic formula worked out line by line. Free, it runs in your browser, and nothing you type leaves your machine. For x² − 5x + 6 = 0 you enter a = 1, b = −5, c = 6 and it answers right away: x = 2 and x = 3.

For homework it's the rounding that bites. Most free solvers hand back a decimal, so when the real answer is 2 + √3 they give you 3.7320508, and if the answer key wanted the exact surd, that decimal is marked wrong. This one keeps the exact form instead of rounding it away on you. When the roots are rational it gives a reduced fraction (2x² + 5x − 3 comes back as x = 1/2 and x = −3, not 0.5), and when they're irrational but tidy it gives a simplified surd like 2 ± √3 with the decimal sitting right beside it. You copy whichever form your teacher asked for.

1. Enter a, the number in front of x². For x² − 5x + 6 = 0 that's 1 (an invisible 1 still counts as 1). It can't be zero. Delete the x² term and you don't have a quadratic anymore.
2. Enter b, the coefficient on the plain x, sign and all. In x² − 5x + 6 that's −5, not 5. This is where most wrong answers come from, a dropped minus sign.
3. Enter c, the constant on its own, with its sign, here 6. If a term is missing from the equation its coefficient is just 0, not a blank box.
4. Read the roots at the top. You get an exact fraction or surd where the numbers are clean, and a decimal where they aren't.
5. Glance at the discriminant and vertex underneath if you need them. The discriminant tells you what kind of roots you got, and the vertex is the turning point of the curve.
6. Copy the worked steps when you have to show your method. The tool drops your a, b, and c into the quadratic formula one line at a time, so you can reproduce the working, not just the answer.

One thing before you start: the equation has to equal zero first. Staring at x² = 3x − 1? Move everything onto one side, x² − 3x + 1 = 0, and then read off a, b, and c. Skip that and every coefficient is wrong.

When the discriminant b² − 4ac is a perfect square, the roots are rational. The tool computes (−b ± √disc) / 2a, then cancels the common factor top and bottom, so you get 1/2 rather than 0.5000000 and −3 rather than −3.0. Reduced to lowest terms, the way a class wants it written.

When the discriminant isn't a perfect square but is still a tidy whole number, the square root simplifies to a surd. √12 becomes 2√3, and the root reads as something like (3 + 2√3) / 2, reduced by the common factor, with the decimal approximation printed right after it. Exact answer for the homework, a number you can picture for the sanity check.

And when there's nothing clean to keep, it falls back to a plain decimal, which is the honest move. Dressing up 6.25 as a fake fraction would be lying about the value.

Two reasons the exact form is worth caring about. First, answer keys usually want it, and a rounded decimal loses marks even when it's technically close. Second, an exact root factors straight back: x = 1/2 and x = −3 tells you the equation is (2x − 1)(x + 3), which a decimal completely hides. One catch, the exact form only triggers when a, b, and c are whole numbers. Feed it decimals and you'll get a decimal back, because there's no clean fraction hiding inside 6.25 anyway.

The discriminant is that b² − 4ac piece sitting under the square root in the formula, and its sign alone tells you the shape of the answer before you finish any arithmetic. Positive means two separate real roots and the parabola crosses the x-axis twice. Zero means one repeated root, where the curve just touches the axis at its lowest (or highest) point. Negative means no real roots at all.

That last case isn't the tool breaking on you. A negative discriminant means the parabola floats entirely above or below the axis and never crosses it, so the answers step into the complex numbers and come back as a conjugate pair, p ± qi. Take x² + 1 = 0: the discriminant is −4 and the roots are +i and −i. A messier one, x² − 4x + 6.25 = 0, has discriminant −9 and roots 2 ± 1.5i. The two answers are always mirror images, same real part, opposite sign on the imaginary bit. Engineers work with these daily, so it's a real answer, not an error dressed up.

There's a shortcut in all this. If a problem only asks whether real solutions exist, the discriminant settles it, positive or zero means yes, negative means no, and you never run the full formula.

The vertex tags along for free too. Every parabola has a turning point, and its x-coordinate is −b / 2a, which is the exact center of the quadratic formula, sitting right between the two roots (the ± √(...) just steps out symmetrically to each side). The tool reports the vertex point and the axis of symmetry x = −b / 2a on every solve. That's the number you want when the question isn't "what are the roots" but "what's the maximum height" or "at what price does profit peak," because the vertex is the high or low point of the whole curve.

You've got 200 metres of fencing and want a rectangle of exactly 2,400 square metres. What dimensions? Call one side x. The perimeter fixes the other at (100 − x), so x(100 − x) = 2400 rearranges to x² − 100x + 2400 = 0. Punch it in, the roots come out x = 40 and x = 60, so the rectangle is 40 by 60.

Notice both roots are real answers here, they're the two sides of the same rectangle, read two ways. That's the pattern behind loads of real quadratics. Anything involving area lands you here, and so does anything moving under gravity, where height over time is −½gt² + v₀t + h₀. Set the height to zero and the roots are the times the object sits at ground level, with the vertex marking the peak of the arc.