Quadratic Formula Calculator

How is the quadratic formula calculated step by step?

One expression solves any equation that can be written ax² + bx + c = 0, namely x = (−b ± √Δ) / 2a, where Δ is the discriminant b² − 4ac. The three coefficients sit in the equation itself, so x² − 3x + 2 = 0 gives a = 1, b = −3 and c = 2.

Take that same equation as a worked example: Δ = (−3)² − 4 × 1 × 2, which is 9 − 8, so Δ = 1. Its square root is 1, the numerator becomes 3 ± 1 and the denominator is 2 × 1 = 2. The ± then splits that fraction in two, giving x₁ = 2 and x₂ = 1.

An absent term has a coefficient of 0, so x² − 3x = 0 goes in as a = 1, b = −3, c = 0. Anything sitting on the right of the equals sign moves across first, since the formula only works on the standard form with 0 on one side.

What does the discriminant tell you about the roots?

Δ settles the question on its own, well before any division has been done, so the calculator puts it at the top of the results. A positive Δ means √Δ is a real number and the ± pulls the numerator apart into two different values, so the parabola cuts the x-axis twice. A Δ of exactly zero makes √Δ = 0, so both branches of the ± land on the same value while the curve only touches the axis at its vertex.

A negative Δ has no real square root at all, which is why the two answers arrive as a conjugate pair such as −1 + 2i and −1 − 2i. The parabola then stays clear of the axis without ever touching it. That sign check also catches typos, since a textbook exercise that turns complex out of nowhere usually means a minus went missing in b or c.

Why can a not be zero in a quadratic equation?

Setting a = 0 wipes out the x² term. What remains is no longer an equation of the second degree: bx + c = 0 is linear, with a single solution at x = −c/b that needs no discriminant at all. The quadratic formula would divide that equation by 2a = 0, so the calculator blocks the entry and names the reason rather than returning an infinity.

Zeros in the other two boxes are legal, since those terms drop out of the equation. Setting b = 0 leaves ax² + c = 0, as in x² − 2 = 0; setting c = 0 leaves ax² + bx = 0, as in x² − 3x = 0 whose roots are 3 and 0.

When does a quadratic formula calculator show a factored form?

A factored form only appears when the brackets are exact, which takes two conditions at once, whole numbers in a, b and c plus a discriminant that is a perfect square such as 1, 4, 9 or 16. Rounded factors would look convincing while multiplying back to the wrong equation, so the tool prints no approximation.

Both conditions hold for 2x² − 3x + 1 = 0, where Δ = 1 and the leading coefficient moves into the first bracket as (2x - 1)(x - 1). A root of zero drops a bracket, which turns x² − 3x = 0 into x(x - 3). A repeated root becomes a square with the leading coefficient in front, so 2x² − 4x + 2 = 0 comes back as 2(x - 1)², while a negative a keeps its sign outside the brackets, as with −x² + 3x − 2 = 0 giving -(x - 1)(x - 2).

The factored form stays empty for x² − 2 = 0, where Δ = 8 and the roots come out at ±1.41421356, shown to eight decimals and ready to be rounded to whatever precision an exercise asks for. The square root of 8 is irrational, so no pair of whole-number brackets multiplies back to that equation. Complex roots stay unfactored for the same reason.

How can the roots be checked without redoing the formula?

Two identities catch a slip faster than a second pass through the whole calculation. The sum of the roots equals −b/a and the product equals c/a, so x² − 3x + 2 = 0 should give a sum of 3 and a product of 2. Its roots, 2 and 1, add to 3 and multiply to 2.

The vertex gives the other check. It sits at x = −b/2a, which doubles as the axis of symmetry, with the two roots the same distance on either side of it. For x² − 3x + 2 = 0 the vertex lands at (1.5, −0.25), exactly halfway between 1 and 2. That negative height on a parabola opening upward confirms two real roots.