Quadratic Equation Solver With Full Working

How to use it

Write your equation in the form ax² + bx + c = 0 first, with everything on one side, then type the three coefficients. Signs matter: in x² − 5x + 6 = 0 the value of b is −5, not 5. Fractions are accepted as they are written, so 1/2 goes in as 1/2 and is kept exact rather than turned into 0.5.

The method box decides which working is shown first, not what is calculated — every applicable route is shown underneath in any case. Auto picks factorising when the discriminant is a perfect square, the formula when it is not, and completing the square when there are no real roots, which is the order an examiner would work in.

If you ask for factorising on an equation that has no whole-number brackets, the tool says so plainly instead of rounding something into brackets that do not exist.

• a — the number in front of x². It cannot be 0, or the equation is not quadratic.
• b — the number in front of x, carrying its sign.
• c — the constant left over once everything is on one side.

How the calculation works

Everything is done in exact fractions. The coefficients are parsed into rationals and stay that way through the discriminant, the roots, the brackets and the completed square, so no rounding error can creep in. A decimal appears only on the last line, and it is labelled as 3 significant figures.

The discriminant b² − 4ac comes first, because it decides everything that follows. Positive means two real roots; zero means one repeated root and a curve that touches the axis; negative means no real roots. If it is a perfect square, the roots are rational, which is exactly the condition for whole-number brackets to exist — that is why the tool can tell you an equation does not factorise rather than leaving you to guess.

For the surd form, the discriminant is split into its largest square factor and a squarefree remainder, so √8 becomes 2√2 and the roots are written over a single denominator the way a mark scheme writes them: (−3 ± √17)/2. Completing the square uses a(x + b/2a)² + (c − b²/4a), which is where the turning point (−b/2a, c − b²/4a) and the line of symmetry come from.

What it deliberately does not do

It gives no complex roots. When the discriminant is negative, the answer at GCSE, IGCSE and Digital SAT level is that there are no real solutions, and the completed square is shown as the proof: a square cannot be negative, so the whole expression has a minimum above the axis or a maximum below it. A pair of numbers with an i in them would be a wrong answer on every one of those papers.

It does not force a factorisation. Plenty of solvers will hand back brackets with decimals in them, which are not brackets any exam would accept and cannot be checked by expanding. If the discriminant is not a perfect square, this tool says there are no whole-number brackets and moves to the formula.

It does not sketch the curve, and it does not claim to know which method your teacher wants. Auto follows the discriminant, which is the mathematical rule rather than a house style.

Where students lose the marks

The most common loss is not the answer but the form of it. A question that says "leave your answer in surd form" wants (−3 ± √17)/2; 0.562 and 3.56 are the same numbers and score less, because the accuracy mark is for the exact value. The reverse also happens — a question asking for 2 decimal places gets a surd and drops the final mark.

The second is a sign error inside the formula. With b = −5, the numerator opens −(−5) = 5, and b² = 25 whether b is 5 or −5. Writing −5² instead of (−5)² turns 25 into −25 and the discriminant with it, which is why the tool prints −b, b² and 4ac as separate lines before combining them.

The third is the turning point read off the completed square with the wrong sign. In (x + 3)² − 4 the turning point is (−3, −4): the x-coordinate is the number that makes the bracket zero, so its sign flips, while the constant outside carries straight through. Reading it as (3, −4) is one of the most reliably lost marks on the higher paper.