Solve x² − 3x − 5 = 0

x² − 3x − 5 = 0 does not factor, and knowing that before searching is the point of this one. The discriminant is (−3)² − 4(1)(−5) = 9 + 20 = 29. Twenty-nine is not a square number, so the roots are irrational and no pair of whole numbers will produce them. The pairs multiplying to −5 are only (1, −...

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 + √29)/2 or (3 − √29)/2

4.192582… · −1.192582…

Both roots are irrational. The surds above are exact; the decimals below them are rounded.

As typedx^2 - 3x - 5 = 0
Expandedx² − 3x − 5 = 0
Standard formx² − 3x − 5 = 0

EACH ROOT PUT BACK, EXACTLY

(3 + √29)/2 → 0 + 0√29

(3 − √29)/2 → 0 + 0√29

sum 3 against −b/a = 3

product −5 against c/a = −5

DISCRIMINANT

29

positive, so two real roots

AXIS OF SYMMETRY

x = 3/2

the roots sit either side of this

AS DECIMALS

4.1925…, −1.1925…

rounded; the surds above are exact

FACTORS OVER ℚ

no

the roots are irrational

WHAT IS HAPPENING

1Expand both sides and collect

x² − 3x − 5 = 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 − 3x − 5 = 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

(−3)² − 4(1)(−5) = 29

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 irrational roots

x = (3 + √29)/2 and x = (3 − √29)/2

The discriminant is not a square number, so the roots are irrational. The surd is the exact answer and the decimal beside it is a rounding of it.

THE CURVEthe curve crosses the axis twice, and the crossings are the roots−4.47.432.8−12.2filled circles are the roots · the gold circle is the turning point at (1.5, −7.3)

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 axis3/2−b ÷ 2a(3 + √29)/2(3 − √29)/2centre 3/2, step ±√29/2the 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

    Compute the discriminant first: (−3)² − 4(1)(−5) = 9 + 20 = 29. Not a square number, so no factorisation exists.

  2. 2

    Apply the formula: x = (3 ± √29) ÷ 2. The −b becomes +3 because b is −3.

  3. 3

    Leave √29 as it is. Twenty-nine is prime, so there is no square factor to pull out and no simpler form.

  4. 4

    Check with the sum: the two roots add to 3, because the √29 parts cancel. That must equal −b/a = 3.

WORKED EXAMPLE

x² − 3x − 5 = 0 Compute the discriminant before anything else: b² − 4ac = (−3)² − 4(1)(−5) = 9 + 20 = 29 Twenty-nine is not a square number, so the roots are irrational and no factorisation over the whole numbers exists. That is worth confirming by hand: the pairs multiplying to −5 are (1, −5) and (−1, 5), adding to −4 and 4 respectively, and neither is 3. There are no other pairs, so the search is finished. Apply the formula: x = (3 ± √29) ÷ 2 The −b is +3 because b itself is −3. As decimals the roots are about 4.19258 and −1.19258. √29 does not simplify. Twenty-nine is prime, so it has no square factor to take outside the root — unlike √8, which is 2√2, or √12, which is 2√3. Check by sum and product rather than by substituting surds. The roots must add to −b/a = 3: the two halves are (3 + √29)/2 and (3 − √29)/2, and adding them cancels the surds, leaving 6/2 = 3. They must multiply to c/a = −5: the product is (9 − 29)/4 = −20/4 = −5. Both hold, and neither required a decimal.

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 = (3 + √29)/2 and x = (3 − √29)/2, about 4.1926 and −1.1926. The discriminant is 29, which is not a square number, so these roots are irrational and cannot be written as fractions or as finite decimals.

  • ·x = (3 ± √29)/2
  • ·About 4.1926 and −1.1926
  • ·The discriminant is 29
  • ·Not a square number, so the roots are irrational

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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.