The Quadratic Formula
The quadratic formula solves any quadratic equation, including the ones that refuse to factorise. On this page you will learn how to rearrange an equation into the standard form, identify the values of a, b and c with their correct signs, substitute carefully, and give answers correct to 3 significant figures or in exact surd form. Worked examples cover the most common sign traps, and the practice room below generates unlimited auto-marked questions across four difficulty levels: Starter, Builder, Challenger and Master.
What is the Quadratic Formula?
Some quadratic equations factorise neatly. Many do not. The quadratic formula works on every quadratic equation, which is why it is the method to reach for when factorising fails or when a question asks for answers to a number of significant figures or decimal places.
If \(ax^{2} + bx + c = 0\) then
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]The formula is printed on the exam formula sheet, but it is used so often that you should know it by heart.
How to use it
- Rearrange the equation into the standard form \(ax^{2} + bx + c = 0\). Everything on one side, zero on the other.
- Write down the values of \(a\), \(b\) and \(c\), including their signs, before touching the formula.
- Substitute carefully. Put negative values in brackets so the signs survive the arithmetic.
- Work out \(b^{2} - 4ac\) first, then square root it.
- Calculate both answers: one using \(+\), one using \(-\). Round only at the very end.
Worked Examples
💡 Example 1: The basic method
Solve \(x^{2} + 3x - 5 = 0\), giving your answers correct to 3 significant figures.
\(a = 1,\; b = 3,\; c = -5\)
\[ x = \frac{-3 \pm \sqrt{3^{2} - 4(1)(-5)}}{2(1)} \]
\[ x = \frac{-3 \pm \sqrt{29}}{2} \]
\(x = 1.19\) or \(x = -4.19\) (3 s.f.)
What's happening?
Write \(a\), \(b\), \(c\) down first. Here \(c = -5\), so \(-4ac = +20\) and \(b^{2} - 4ac = 9 + 20 = 29\).
Calculate \(\dfrac{-3 + \sqrt{29}}{2}\) and \(\dfrac{-3 - \sqrt{29}}{2}\) separately, then round.
💡 Example 2: When \(b\) is negative
Solve \(2x^{2} - 7x + 4 = 0\), giving your answers correct to 3 significant figures.
\(a = 2,\; b = -7,\; c = 4\)
\[ x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4(2)(4)}}{2(2)} \]
\[ x = \frac{7 \pm \sqrt{17}}{4} \]
\(x = 2.78\) or \(x = 0.719\) (3 s.f.)
What's happening?
\(-b\) becomes \(-(-7) = +7\), and \((-7)^{2} = 49\) is positive. This is where most formula marks are lost.
\(b^{2} - 4ac = 49 - 32 = 17\).
💡 Example 3: Rearrange first
Solve \(3x^{2} + 4x = 2\), giving your answers correct to 3 significant figures.
Rearrange: \(3x^{2} + 4x - 2 = 0\)
\(a = 3,\; b = 4,\; c = -2\)
\[ x = \frac{-4 \pm \sqrt{16 + 24}}{6} = \frac{-4 \pm \sqrt{40}}{6} \]
\(x = 0.387\) or \(x = -1.72\) (3 s.f.)
What's happening?
The 2 must be brought over to the left before reading off the values, so \(c = -2\), not \(2\).
Substituting from the original form is the classic error here.
💡 Example 4: Surd form
Solve \(x^{2} + 10x + 3 = 0\), giving your answers in surd form.
\(a = 1,\; b = 10,\; c = 3\)
\[ x = \frac{-10 \pm \sqrt{100 - 12}}{2} = \frac{-10 \pm \sqrt{88}}{2} \]
\[ x = \frac{-10 \pm 2\sqrt{22}}{2} \]
\(x = -5 + \sqrt{22}\) or \(x = -5 - \sqrt{22}\)
What's happening?
"Surd form" or "exact answers" means no rounding at all.
Simplify the surd first: \(\sqrt{88} = \sqrt{4 \times 22} = 2\sqrt{22}\). Then every term in the numerator divides by 2.
🔑 Key Points
- The formula solves any quadratic, so it is the safe fallback when factorising fails.
- \(a\) is the coefficient of \(x^{2}\); it is not necessarily the first number written in the equation.
- If a question asks for answers to a number of s.f. or d.p., the quadratic formula is almost certainly required.
- If \(b^{2} - 4ac\) is a perfect square, the quadratic would also factorise; the formula still works.
- For surd-form answers, simplify \(\sqrt{b^{2} - 4ac}\) before dividing.
⚠️ Common Pitfalls
- Sign errors when \(b\) is negative: \(-b\) becomes positive and \(b^{2}\) is always positive.
- Forgetting that the whole numerator, \(-b \pm \sqrt{b^{2}-4ac}\), is divided by \(2a\).
- Substituting before rearranging into \(ax^{2} + bx + c = 0\).
- Rounding \(\sqrt{b^{2}-4ac}\) mid-calculation, which loses accuracy marks.
- Getting a negative under the square root: recheck \(a\), \(b\) and \(c\) before concluding there are no solutions.