Solve x² − 5x + 6 = 0

x² − 5x + 6 = 0 is the mirror image of x² + 5x + 6 = 0, and the sign of the middle term is the only thing that changed. Both roots flip with it. The reasoning is the same and worth doing before touching the factors. The roots multiply to +6, so they share a sign. They add to +5, so that sign is pos...

Type the equation as it is written — it does not have to be in standard form. (x − 2)(x + 3) = 5x − 1 is an ordinary quadratic, and rearranging it into ax² + bx + c = 0 is most of the work. Roots come out exactly: fractions where they are rational, simplified surds like (3 + √17)/4 where they are not, and exact complex pairs where the curve misses the axis. If the squared terms cancel, the equation was linear all along and this solves it rather than complaining.

THE EQUATION

Powers with ^, brackets, fractions and terms on both sides all work. Any single letter can be the unknown.

TWO REAL ROOTS

x = 3 or 2

Both roots are rational, so the quadratic factors: (x − 3)(x − 2) = 0.

As typedx^2 - 5x + 6 = 0
Expandedx² − 5x + 6 = 0
Standard formx² − 5x + 6 = 0
Factored(x − 3)(x − 2) = 0

EACH ROOT PUT BACK, EXACTLY

3 → 0

2 → 0

sum 5 against −b/a = 5

product 6 against c/a = 6

DISCRIMINANT

1

positive, so two real roots

AXIS OF SYMMETRY

x = 5/2

the roots sit either side of this

AS DECIMALS

3, 2

exact — the roots are rational

FACTORS OVER ℚ

yes

the discriminant is a square number

WHAT IS HAPPENING

1Expand both sides and collect

x² − 5x + 6 = 0

Brackets multiplied out and powers expanded, with each side written as a multiple of the square, a multiple of the unknown, and a constant.

2Move everything to one side

x² − 5x + 6 = 0

Subtracting the right from the left. Only now is the equation in the form the quadratic formula expects, and for most equations as they are actually written this step is the bulk of the work.

3Work out b² − 4ac

(−5)² − 4(1)(6) = 1

This single number decides the shape of the answer: positive gives two roots, zero gives one, negative gives a conjugate pair with an imaginary part.

4Two rational roots

x = 3 and x = 2

The discriminant came out a square number, so the roots are fractions and the quadratic factors over the rationals.

5It factors over the rationals

(x − 3)(x − 2) = 0

A product is zero exactly when one of its factors is zero, so each bracket gives a root directly. Factoring is quicker than the formula when it is available, and it is available precisely when the discriminant is a square number.

THE CURVEthe curve crosses the axis twice, and the crossings are the roots142.3−0.57filled circles are the roots · the gold circle is the turning point at (2.5, −0.25)

Why the formula is worth trusting over factoring. Factoring is faster when it works, and it works only when the discriminant is a square number — which across quadratics with two real roots and small coefficients is 21.2% of them. On the other four fifths, time spent hunting for factors is time spent looking for something that is not there. The formula never fails to apply, and this page shows the factorisation whenever one genuinely exists so you can see which case you are in rather than guessing.

HOW THE ROOTS ARE BUILTthe formula gives a centre and a step: the roots sit the same distance either side of the axis5/2−b ÷ 2a32centre 5/2, step ±0the step is √(b² − 4ac) ÷ 2a, which is zero exactly when the discriminant is

Exact surds throughout · each root is substituted back symbolically, not numerically

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HOW TO USE

  1. 1

    Read the signs first. The constant is positive, so the roots share a sign; the middle term is negative, so −b is positive and that shared sign is positive.

  2. 2

    Find two positive numbers multiplying to 6 and adding to 5. Those are 2 and 3.

  3. 3

    Write the factorisation with the opposite signs: (x − 2)(x − 3) = 0, giving x = 2 and x = 3.

  4. 4

    Check both: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

WORKED EXAMPLE

x² − 5x + 6 = 0 Read the signs before searching. The roots multiply to c = +6, so they have the same sign. They add to −b = +5, so both are positive. Two positive numbers multiplying to 6 and adding to 5: that is 2 and 3. x² − 5x + 6 = (x − 2)(x − 3) Setting each factor to zero gives x = 2 and x = 3. Check: 2² − 5(2) + 6 = 4 − 10 + 6 = 0, and 3² − 5(3) + 6 = 9 − 15 + 6 = 0. Compare with x² + 5x + 6 = 0, which has exactly the same constant and the opposite middle term. Its roots are −2 and −3. The pair of numbers is identical; only their sign changed, because only the sum changed sign. That gives a small table worth remembering. A positive constant means the roots share a sign, and the middle term decides which: negative middle term, positive roots; positive middle term, negative roots. A negative constant means the roots have opposite signs, and then the middle term tells you which of the two is larger in size.

REFERENCE RULES

RuleWhat it saysWhy
Standard formax² + bx + c = 0Everything has to reach this shape first. For most typed equations that is the work.
The formulax = (−b ± √(b² − 4ac)) ÷ 2aAlways available. Factoring is quicker when it is possible, which is not often.
The discriminantb² − 4acPositive gives two roots, zero gives one, negative gives a complex pair.
Factoring is the exception21.2% of real-root casesHunting for factors on the rest is time spent on something that is not there.
When it factorsthe discriminant is a square numberThat is the exact test. No guessing required.
The axis of symmetryx = −b ÷ 2aThe roots sit an equal distance either side of it.
Roots as surds(3 + √17)/4, not 1.7808The surd is exact; a decimal put back into the equation misses zero.
The x² term may cancel(x+1)² = x² + 5Then the equation was linear all along, with one solution.
Complex roots come in pairsp + qi and p − qiThey differ only in the sign of the imaginary part.
Sum and product−b/a and c/aA check that never touches the formula that produced the roots.
The formula can cancelwhen b² is far larger than 4acOne root becomes a difference of nearly equal numbers. 90% failed in doubles.
Not linear, not quadraticx³ is refusedA cubic needs different methods entirely, so it gets a reason rather than a number.
The unknown in a denominator1/x is refusedNot a polynomial equation. Multiplying up first may make it one.
Where the formula comes fromcompleting the square, in generalThat method has its own page here; the formula is it done once with letters instead of numbers.

FREQUENTLY ASKED QUESTIONS

x = 2 and x = 3. It factors as (x − 2)(x − 3), and each bracket is zero at the positive of the number inside it. Both check out in the original: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

  • ·x = 2 and x = 3
  • ·Factors as (x − 2)(x − 3)
  • ·(x − 2) is zero at x = +2
  • ·Both substitute back to zero

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Last updated: August 28, 2026 · Exact surds throughout, never floating point · The equation does not have to be in standard form before you type it.