Which method for solving a quadratic is fastest?

Factoring, when it works — but it only works when b² − 4ac is a perfect square. Evaluate the discriminant first: one multiplication and one subtraction tells you whether factoring is worth attempting at all, and if it is not, the quadratic formula is the fastest remaining route.

Do the three methods ever give different answers?

No. They are three routes to the same two roots, and any disagreement is an arithmetic slip. The quickest check is Vieta's: the roots must add to −b/a and multiply to c/a, whichever method produced them.

When should I complete the square instead of using the formula?

Whenever the question is about the curve rather than its roots — a maximum, a minimum, the vertex, or a graph to sketch by transformation. Completing the square produces a(x − h)² + k, which puts the vertex at (h, k) directly; the formula gives the roots and nothing about where the curve turns.

What does it mean when the discriminant is negative?

There are no real roots, so the parabola never touches the x-axis. The two solutions are a complex conjugate pair, which the quadratic formula returns normally once you allow the square root of a negative number. Calling it “no solution” is only accurate if you have agreed in advance to stay inside the real numbers.

Can every quadratic be factored?

Over the integers, no — only those whose discriminant is a perfect square. Over the real numbers, any quadratic with a non-negative discriminant factors, but the factors contain surds: x² − 2 is (x − √2)(x + √2). Over the complex numbers every quadratic factors. Which is meant depends on the question, and school questions almost always mean over the integers.

Which Method to Use to Solve a Quadratic

The one number that decides the method

Before picking a method, evaluate b² − 4ac. Its sign and its shape between them tell you almost everything about what is coming. If it is a positive perfect square — 1, 4, 9, 16, 25 — the roots are rational, the quadratic factors over the integers, and factoring is by far the fastest route. If it is positive but not a perfect square, the roots are real and irrational: no integer factorisation exists, and time spent hunting for one is time wasted. If it is zero, the trinomial is a perfect square and the single repeated root is −b/2a. If it is negative, there are no real roots at all and the answer is a complex conjugate pair.

That check costs one multiplication and one subtraction, and it is worth doing every single time. The most common way to lose minutes on a quadratic is to try factoring an equation that has no integer factorisation — there is no signal that tells you to stop, so you keep trying pairs. Evaluating the discriminant first turns that open-ended search into a yes-or-no question you can answer in seconds.

Δ = b² − 4ac

Factoring: fastest when it works, useless when it does not

Factoring rewrites ax² + bx + c as a product of two brackets, and the roots then fall out of the zero-product rule: if two things multiply to zero, one of them is zero. The method is the AC search — find two integers that multiply to a×c and add to b, split the middle term with them, and group. For x² − 5x + 6 the pair is −2 and −3, the factorisation is (x − 2)(x − 3), and the roots are 2 and 3 with no further work at all.

Two habits make it reliable. Take any common factor out first: 2x² + 10x + 12 is 2(x + 2)(x + 3), and an answer of (2x + 4)(x + 3) is marked wrong because the first bracket is still divisible. And check the discriminant before starting, because the whole method depends on that pair of integers existing, which happens precisely when b² − 4ac is a perfect square.

Worked example

x² − 5x + 6: a×c = 6 and b = −5, so the pair is −2 and −3. The factorisation is (x − 2)(x − 3) and the roots are 2 and 3.

Completing the square: slower, but it gives you the vertex

Completing the square rewrites the quadratic as a(x − h)² + k. Solving from there is straightforward — isolate the squared bracket, take the square root of both sides keeping the plus-or-minus, and move h across — but the roots are not really the point. The point is h and k, which are the coordinates of the vertex, and no other method produces them as a by-product.

So the method earns its place whenever the question is about the curve rather than about where it crosses the axis: a maximum, a minimum, a graph to sketch by transformation, or a quadratic inside an integral that needs to become a standard form. The step people drop is the subtraction — you must add AND subtract (b/2a)², because adding alone changes the expression rather than rewriting it. Multiplying the answer back out catches that instantly.

a(x − h)² + k, with h = −b/2a and k = c − b²/4a

The quadratic formula: always works, and where its signs go wrong

The formula is what you get by completing the square on ax² + bx + c in general rather than on one particular equation, which is why the 2a underneath is the a you divided by and why b² − 4ac sits under the radical. It solves every quadratic, factorable or not, and it is the right default when the discriminant is not a perfect square or when the coefficients are awkward.

Almost every wrong answer from the formula is a sign error rather than an algebra error, and there are exactly two places it happens. First, −b when b is already negative: in x² − 7x + 12, −b is +7, not −7. Second, squaring a negative b: (−7)² is +49, and typing -7^2 into most calculators gives −49 because the exponent binds tighter than the minus. Writing the substitution out with brackets before evaluating anything is the habit that prevents both.

Whichever method you use, the answer can be checked in seconds by Vieta's relations: the two roots must add to −b/a and multiply to c/a. If either fails, a sign went astray somewhere above.

x = (−b ± √(b² − 4ac)) / 2a

The same equation, all three ways

Take 2x² + 5x − 3 = 0. Its discriminant is 25 − 4(2)(−3) = 25 + 24 = 49, which is 7², so it factors and all three methods will work.

Factoring: a×c = −6 and the pair is 6 and −1, giving 2x² + 6x − x − 3, then 2x(x + 3) − 1(x + 3), then (x + 3)(2x − 1). The roots are −3 and 1/2. Completing the square: 2(x + 5/4)² − 49/8 = 0, so (x + 5/4)² = 49/16, so x + 5/4 = ±7/4, giving the same two roots and telling you the vertex sits at (−5/4, −49/8). The formula: x = (−5 ± 7)/4, which is −3 and 1/2. Three routes, one answer, and only the middle one told you where the curve turns.

Worked example

2x² + 5x − 3 = 0 → Δ = 49 → (x + 3)(2x − 1) → x = −3 or x = 1/2, vertex at (−5/4, −49/8).

When the answer is not a decimal

A large share of quadratics have irrational roots, and for those the exact form is the answer while the decimal is a rounding of it. x² − 2 = 0 has roots ±√2; writing 1.4142136 is not a simplification of that, it is a loss. Marking schemes ask for the surd, and so does any further algebra you do with the result.

Getting the exact form right takes two separate pieces of care. The radical has to be simplified rather than left as it fell out — √8 is 2√2 — and the fraction has to reduce, cancelling any common factor between −b, the surd, and 2a. A calculator that reports only decimals has skipped both.

A short decision rule

Evaluate b² − 4ac. If it is a perfect square, factor. If the question asks for a maximum, a minimum, a vertex, or a graph, complete the square regardless of what the discriminant says. Otherwise use the formula. If the discriminant is negative, the formula is the only one of the three that returns anything, and what it returns is a complex conjugate pair rather than nothing.

One case is not a quadratic at all: if a is zero the x² term is gone, the equation is linear, and dividing by 2a would be dividing by zero. Its single root is −c/b.

Common mistakes

Trying to factor before checking the discriminant. If b² − 4ac is not a perfect square there is no integer factorisation to find, and nothing in the search itself tells you to stop looking.
Reading −b as negative when b is already negative. In x² − 7x + 12, b is −7 and −b is +7; getting this wrong flips both roots and produces an answer plausible enough to hand in.
Squaring a negative b and keeping the minus. (−7)² is +49. Typed as -7^2 most calculators return −49, because the exponent binds more tightly than the minus sign.
Adding the completing term without subtracting it. Adding (b/2a)² alone changes the expression into a different one; the subtraction is what makes it a rewrite rather than a new problem.
Leaving a common factor inside the brackets. (2x + 4)(x + 3) multiplies out correctly and is still marked wrong, because 2x + 4 is not fully factored.
Reporting a decimal where the root is irrational. √2 is the answer; 1.4142136 is a rounding of it, and it cannot be un-rounded later.

When not to rely only on the calculator

These three methods solve quadratic equations in one variable, and nothing else. A cubic or higher needs a different approach entirely, and a quadratic in two variables — an ellipse, a hyperbola — is a different subject. The algebra also says nothing about whether a root makes sense in the situation being modelled: a projectile problem will happily produce a negative time, and a geometry problem a negative length, and discarding those is a modelling judgement rather than an algebraic one. Where a quadratic is standing in for something physical, check both roots against the situation before using either.

Factoring, Completing the Square, or the Quadratic Formula — Which to Use