2x² + 5x − 3 = 0 is the first shape where the easy factoring trick stops working. With a leading coefficient of 1 you look for two numbers that multiply to the constant and add to the middle coefficient. With a 2 in front, that shortcut gives the wrong pair. The version that does work multiplies th...
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 = 1/2 or −3
Both roots are rational, so the quadratic factors: (2x − 1)(x + 3) = 0.
| As typed | 2x^2 + 5x - 3 = 0 |
| Expanded | 2x² + 5x − 3 = 0 |
| Standard form | 2x² + 5x − 3 = 0 |
| Factored | (2x − 1)(x + 3) = 0 |
EACH ROOT PUT BACK, EXACTLY
1/2 → 0 ✓
−3 → 0 ✓
sum −5/2 against −b/a = −5/2 ✓
product −3/2 against c/a = −3/2 ✓
DISCRIMINANT
49
positive, so two real roots
AXIS OF SYMMETRY
x = −5/4
the roots sit either side of this
AS DECIMALS
0.5, −3
exact — the roots are rational
FACTORS OVER ℚ
yes
the discriminant is a square number
WHAT IS HAPPENING
2x² + 5x − 3 = 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.
2x² + 5x − 3 = 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(2)(−3) = 49
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 = 1/2 and x = −3
The discriminant came out a square number, so the roots are fractions and the quadratic factors over the rationals.
(2x − 1)(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
Multiply the outer coefficients: a × c = 2 × (−3) = −6. This is the number the pair must multiply to, not the constant alone.
Find two numbers multiplying to −6 and adding to 5. Those are 6 and −1.
Split the middle term and factor in pairs: 2x² + 6x − x − 3 becomes 2x(x + 3) − 1(x + 3), so (2x − 1)(x + 3).
Set each bracket to zero: 2x − 1 = 0 gives x = 1/2, and x + 3 = 0 gives x = −3. Check both in the original.
2x² + 5x − 3 = 0 The shortcut for a leading coefficient of 1 does not apply, so use the version that does. Multiply the outer coefficients: 2 × (−3) = −6. Find two numbers multiplying to −6 and adding to 5. The pairs for −6 are (1, −6), (−1, 6), (2, −3) and (−2, 3); the one adding to 5 is 6 and −1. Split the middle term using them: 2x² + 6x − x − 3 Factor in pairs: 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3) Setting each factor to zero: 2x − 1 = 0 → x = 1/2 x + 3 = 0 → x = −3 Check both in the original. At x = 1/2: 2(1/4) + 5(1/2) − 3 = 1/2 + 5/2 − 3 = 3 − 3 = 0. At x = −3: 2(9) + 5(−3) − 3 = 18 − 15 − 3 = 0. By formula, for comparison: the discriminant is 5² − 4(2)(−3) = 25 + 24 = 49, which is a square number — that is exactly why a factorisation existed. The roots are (−5 ± 7) ÷ 4, giving 2/4 = 1/2 and −12/4 = −3. The fractional root is worth expecting rather than being surprised by. A bracket like (2x − 1) is zero at x = 1/2, and a leading coefficient other than 1 puts numbers in front of the x inside the brackets, which is precisely what produces fractions.
| 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 = 1/2 and x = −3. It factors as (2x − 1)(x + 3), so the brackets give 1/2 and −3. Both check out: at x = 1/2 the expression is 1/2 + 5/2 − 3 = 0, and at x = −3 it is 18 − 15 − 3 = 0.
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Last updated: August 31, 2026 · Exact surds throughout, never floating point · The equation does not have to be in standard form before you type it.