x² + 5x + 6 = 0 factors cleanly, and it is a good place to notice something that saves time on every quadratic afterwards: every coefficient is positive and both roots are negative. That is not a coincidence. The two roots multiply to c, which is +6, so they share a sign. They add to −b, which is −...
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.
Powers with ^, brackets, fractions and terms on both sides all work. Any single letter can be the unknown.
TWO REAL ROOTS
x = −2 or −3
Both roots are rational, so the quadratic factors: (x + 2)(x + 3) = 0.
| As typed | x^2 + 5x + 6 = 0 |
| Expanded | x² + 5x + 6 = 0 |
| Standard form | x² + 5x + 6 = 0 |
| Factored | (x + 2)(x + 3) = 0 |
EACH ROOT PUT BACK, EXACTLY
−2 → 0 ✓
−3 → 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
−2, −3
exact — the roots are rational
FACTORS OVER ℚ
yes
the discriminant is a square number
WHAT IS HAPPENING
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.
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.
(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.
x = −2 and x = −3
The discriminant came out a square number, so the roots are fractions and the quadratic factors over the rationals.
(x + 2)(x + 3) = 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.
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.
Exact surds throughout · each root is substituted back symbolically, not numerically
Look for two numbers that multiply to 6 and add to 5. Because both coefficients are positive, both numbers will be positive and both roots will be negative.
Those numbers are 2 and 3, so the factorisation is (x + 2)(x + 3) = 0.
Set each bracket to zero: x + 2 = 0 gives x = −2, and x + 3 = 0 gives x = −3. The sign flips as the term moves across.
Check both in the original: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.
x² + 5x + 6 = 0 Find two numbers that multiply to 6 and add to 5. The pairs multiplying to 6 are 1 and 6, and 2 and 3. The second pair adds to 5, so: x² + 5x + 6 = (x + 2)(x + 3) Setting each factor to zero: x + 2 = 0 → x = −2 x + 3 = 0 → x = −3 Check: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0, and (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0. Why both roots are negative, before doing any of that. The roots of ax² + bx + c multiply to c/a and add to −b/a. Here they multiply to +6, so they have the same sign, and they add to −5, so that sign is negative. All-positive coefficients always give roots that are negative or complex — worth knowing, because it cuts the search in half before it starts. And the sign flip in the brackets. (x + 2) is zero when x is −2, because that is what makes the bracket vanish. Reading +2 straight off is the single most common error here, and it gives an answer that fails the check immediately: 4 + 10 + 6 = 20, not 0.
| Rule | What it says | Why |
|---|---|---|
| Standard form | ax² + bx + c = 0 | Everything has to reach this shape first. For most typed equations that is the work. |
| The formula | x = (−b ± √(b² − 4ac)) ÷ 2a | Always available. Factoring is quicker when it is possible, which is not often. |
| The discriminant | b² − 4ac | Positive gives two roots, zero gives one, negative gives a complex pair. |
| Factoring is the exception | 21.2% of real-root cases | Hunting for factors on the rest is time spent on something that is not there. |
| When it factors | the discriminant is a square number | That is the exact test. No guessing required. |
| The axis of symmetry | x = −b ÷ 2a | The roots sit an equal distance either side of it. |
| Roots as surds | (3 + √17)/4, not 1.7808 | The surd is exact; a decimal put back into the equation misses zero. |
| The x² term may cancel | (x+1)² = x² + 5 | Then the equation was linear all along, with one solution. |
| Complex roots come in pairs | p + qi and p − qi | They differ only in the sign of the imaginary part. |
| Sum and product | −b/a and c/a | A check that never touches the formula that produced the roots. |
| The formula can cancel | when b² is far larger than 4ac | One root becomes a difference of nearly equal numbers. 90% failed in doubles. |
| Not linear, not quadratic | x³ is refused | A cubic needs different methods entirely, so it gets a reason rather than a number. |
| The unknown in a denominator | 1/x is refused | Not a polynomial equation. Multiplying up first may make it one. |
| Where the formula comes from | completing the square, in general | That method has its own page here; the formula is it done once with letters instead of numbers. |
x = −2 and x = −3. The expression factors as (x + 2)(x + 3), and each bracket is zero at the negative of the number inside it. Substituting confirms both: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.
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Last updated: September 1, 2026 · Exact surds throughout, never floating point · The equation does not have to be in standard form before you type it.