How to solve quadratic equations — three methods, honestly compared

Every quadratic ax² + bx + c = 0 yields to the same three methods. This guide works one example per method — fully, with the verification most guides skip — and tells you when each method is actually the right tool.

Choosing a method in five seconds

Factoring

Try first when the discriminant is a perfect square — or on sight, for small friendly numbers.

Fastest when it works

Quadratic formula

Everything else. Ugly coefficients, no factor pair, non-perfect-square discriminant, complex roots.

Always works

Completing the square

When you need vertex form, or the question explicitly asks for it. Also the proof behind the formula.

Slowest — most instructive

Step zero, always: get standard form first — everything on one side, zero on the other. x² = 3x becomes x² − 3x = 0 before any method applies.

Method 1 — Factoring

If the quadratic splits into two brackets, the zero product property finishes the job: a product is zero only when one of its factors is. The skill is finding the pair of numbers that multiply to c and add to b.

The pattern

x^2 + bx + c = (x + p)(x + q), \quad pq = c, \ p + q = b

Solve x² + 5x + 6 = 0

1.Find two numbers with product 6 and sum 5: 2 and 3
2.Factor: (x + 2)(x + 3) = 0
3.Zero product property: x + 2 = 0 or x + 3 = 0

x = −2 or x = −3

Verified: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓ and (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0 ✓.

Method 2 — The quadratic formula

The universal solvent. Identify a, b and c (with their signs), compute the discriminant, and substitute carefully. The discriminant tells you in advance what kind of roots to expect.

The formula

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Solve 2x² − 7x + 3 = 0

1.Identify: a = 2, b = −7, c = 3
2.Discriminant: (−7)² − 4(2)(3) = 49 − 24 = 25 — a perfect square, so the roots are rational
3.x = (7 ± √25) / 4 = (7 ± 5) / 4

x = 3 or x = 1/2

Verified: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0 ✓ and 2(1/4) − 7(1/2) + 3 = 0.5 − 3.5 + 3 = 0 ✓.

Method 3 — Completing the square

Force the left side into a perfect square by adding (b/2)² to both sides. Slower for solving — but it converts the equation to vertex form and it is the derivation of the quadratic formula itself.

The key move

x^2 + bx \;\longrightarrow\; \left(x + \dfrac{b}{2}\right)^2 \quad \text{by adding } \left(\dfrac{b}{2}\right)^2

Solve x² + 6x + 5 = 0

1.Move the constant: x² + 6x = −5
2.Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4
3.Write as a square: (x + 3)² = 4
4.Take roots (with ±): x + 3 = 2 or x + 3 = −2

x = −1 or x = −5

Verified: 1 − 6 + 5 = 0 ✓ and 25 − 30 + 5 = 0 ✓.

Now practise — with a solver that checks itself

Try each method by hand first, then run the same problem here. The solver factors when it can and uses the formula when it must — and plugs every root back in, the habit this guide keeps nagging you about.

What are we solving?

Tip: use ; between two equations for a system, ^ for powers, / for fractions.

The three errors that cost the most marks

Reading b without its sign

In x² − 5x + 6, b = −5 and −b = +5. Write a, b, c down with signs before substituting — every worked example above does, on purpose.

Losing the second root

The ± is not decoration. x + 3 = ±2 is two equations, and x² = 9 is x = 3 AND x = −3. If your answer has one root and the discriminant was positive, you dropped one.

Skipping the plug-back check

Substituting your roots into the original takes seconds and catches sign slips, arithmetic slips and dropped roots in one move. It is the entire premise of this site — make it your premise too.

Solving quadratics — questions

Which method should I use first?

Try factoring — if the quadratic factors over the integers, it is the fastest route by far. If no factor pair appears within a few seconds of looking, switch to the quadratic formula; it always works. Completing the square is rarely the fastest for solving, but it is the method that produces vertex form, and it is how the formula itself is derived.

How do I know if a quadratic will factor?

Compute the discriminant b² − 4ac. If it is a perfect square (0, 1, 4, 9, 16, 25…), the roots are rational and the quadratic factors over the rationals. If it is not a perfect square, do not waste time hunting for a factor pair that does not exist — go straight to the formula.

Why do quadratics have two solutions?

Because a parabola crosses the x-axis in up to two places, and the equation asks where y = 0. The two crossings are the two roots. When the discriminant is zero the parabola just touches the axis (one repeated root); when it is negative the parabola never reaches the axis and the roots are complex numbers.

What is the most common mistake when solving quadratics?

Sign errors with b in the formula — in x² − 5x + 6, b is −5. A close second: forgetting the ±, which throws away one of the two roots. Writing out a, b, c explicitly before substituting prevents both. And whatever method you used, plug the roots back in: ten seconds, total certainty.

Do I always need standard form first?

Yes. Every method assumes ax² + bx + c = 0 with everything on one side. x² = 3x is not ready until it is x² − 3x = 0. Moving terms across and flipping their signs is step zero of every worked example on this page.