What Is the Quadratic Formula?
The quadratic formula is a universal tool for solving any equation of the form ax² + bx + c = 0 where a ≠ 0. The formula gives both solutions directly:
x = (−b ± √(b² − 4ac)) / 2aThe beauty of the quadratic formula is that it works every single time. Factoring requires you to spot integer pairs that multiply to ac and add to b — possible for some equations, frustrating for others. Completing the square always works but takes several careful steps. The quadratic formula collapses all of that into one plug-and-chug expression.
The ± symbol indicates two solutions: one using the plus sign and one using the minus sign. These are called the roots or zeros of the quadratic, and geometrically they are the x-intercepts of the parabola y = ax² + bx + c. Every standard-form quadratic produces zero, one, or two real roots depending on the sign of the expression under the radical (more on that in the discriminant section).
The formula was known in essentially its modern form by mathematicians of the Islamic Golden Age in the 9th century, and the general approach — completing the square — appears in Babylonian tablets dating back nearly 4,000 years. Today it remains one of the most important formulas a math student ever memorizes, appearing in physics (projectile motion), engineering (vibration analysis), economics (revenue optimization), and computer science (root-finding algorithms).
Deriving the Quadratic Formula from Completing the Square
The quadratic formula isn't magic — it comes from completing the square applied to the general form. Understanding the derivation helps you remember the formula and recognize when it applies. Start with ax² + bx + c = 0.
- Divide by a
x² + (b/a)x + (c/a) = 0. We need a leading coefficient of 1 so completing the square works cleanly. - Move the constant
x² + (b/a)x = −c/a. - Complete the square on the left
Take half of the x-coefficient:
b/(2a). Square it:b²/(4a²). Add this to both sides:x² + (b/a)x + b²/(4a²) = −c/a + b²/(4a²). - Factor the left side as a perfect square
(x + b/(2a))² = b²/(4a²) − c/a. - Combine the right side over a common denominator
(x + b/(2a))² = (b² − 4ac) / (4a²). - Take the square root of both sides
x + b/(2a) = ±√(b² − 4ac) / (2a). - Solve for x
x = −b/(2a) ± √(b² − 4ac) / (2a), which simplifies tox = (−b ± √(b² − 4ac)) / (2a). That's the quadratic formula.
Every term has a clear origin: the −b/(2a) is the x-coordinate of the vertex (the axis of symmetry), and the ±√(b² − 4ac) / (2a) is how far the two roots sit on either side of that axis.
Using the Quadratic Formula Step by Step
Applying the quadratic formula is straightforward once the equation is in standard form.
- Write the equation in the form
ax² + bx + c = 0Move every term to one side so the right side is zero. If terms are in a different order, rearrange them.
- Identify a, b, and c
These are the coefficients including their signs. If a term is missing (for example
x² − 9 = 0), the missing coefficient is zero (b = 0). - Substitute into the formula
Use parentheses around every coefficient to avoid sign mistakes:
x = (−(b) ± √((b)² − 4(a)(c))) / (2(a)). - Simplify the discriminant
b² − 4acfirstCalculate this single number before doing anything else; it tells you immediately what kind of answer to expect.
- Take the square root
Simplify the radical if possible (factor out perfect squares).
- Compute both solutions
Use the
+sign for one root and the−sign for the other. - Check your answers
Substitute back into the original equation.
2x² − 7x + 3 = 0Here a = 2, b = −7, c = 3.
Discriminant: (−7)² − 4(2)(3) = 49 − 24 = 25.
Square root: √25 = 5.
Solutions: x = (7 ± 5)/4. So x = 12/4 = 3 or x = 2/4 = 1/2.
Check: 2(3)² − 7(3) + 3 = 18 − 21 + 3 = 0 ✓ and 2(1/2)² − 7(1/2) + 3 = 1/2 − 7/2 + 3 = 0 ✓.
x = 3 or x = 1/2The Discriminant — What b² − 4ac Tells You
The discriminant is the expression under the radical: Δ = b² − 4ac. Its sign tells you everything about the nature of the solutions before you even finish the formula.
| Case | Condition | Result | Why |
|---|---|---|---|
| 1 | Δ > 0 | Two distinct real solutions | The parabola crosses the x-axis twice. If Δ is also a perfect square, the solutions are rational and the quadratic factors over the integers. If not, the solutions are irrational and involve a radical. |
| 2 | Δ = 0 | One repeated real solution | Sometimes called a double root. The parabola just touches the x-axis at its vertex. The single root is x = −b/(2a). |
| 3 | Δ < 0 | No real solutions | The parabola doesn't cross the x-axis at all. The solutions are complex conjugates of the form x = (−b ± i√|Δ|) / (2a). |
For x² − 5x + 6 = 0, discriminant = 25 − 24 = 1 > 0 (perfect square), so two rational roots.
For x² − 6x + 9 = 0, discriminant = 36 − 36 = 0, so one repeated root. The quadratic factors as (x − 3)² = 0, giving x = 3 (multiplicity 2).
For x² + 2x + 5 = 0, discriminant = 4 − 20 = −16 < 0, so two complex solutions: x = (−2 ± 4i)/2 = −1 ± 2i.
Checking the discriminant first is one of the highest-leverage habits in algebra. It tells you whether to expect clean numbers, irrational answers, or complex ones, so you can verify your work as you go.
Quadratic Formula vs Factoring vs Completing the Square
Three methods can solve a quadratic equation, and choosing the right one saves time on exams.
Factoring — fastest when it works
Factoring is fastest when it works. If you can quickly spot two numbers that multiply to ac and add to b, write the factored form and read off the roots. Factoring works only when the discriminant is a non-negative perfect square — otherwise the roots aren't rational and you'll waste time hunting for factors that don't exist.
Completing the square — for structure and conversion
Completing the square is essential when you need to convert ax² + bx + c into vertex form a(x − h)² + k — for graphing, finding the maximum or minimum, or deriving the quadratic formula itself. It's also used in calculus when integrating expressions involving x² + bx + c under a radical.
The quadratic formula — universal fallback
The quadratic formula is the universal fallback. Use it when factoring fails, when the discriminant isn't a perfect square, or when you need exact (potentially irrational or complex) roots.
First check the discriminant. If it's a small perfect square, try factoring. If the leading coefficient is 1 and the linear coefficient is even, completing the square gives clean intermediate steps. Otherwise, go straight to the quadratic formula.
x² + 6x + 5 = 0By factoring: (x + 1)(x + 5) = 0, so x = −1 or x = −5.
By the quadratic formula: discriminant = 36 − 20 = 16, x = (−6 ± 4)/2 = −1, −5.
Same answer, but factoring took two steps and the formula took four.
x = −1 or x = −5x² + 6x − 4 = 0Factoring fails (no integer pair works), so go straight to the formula: discriminant = 36 + 16 = 52, x = (−6 ± √52)/2 = −3 ± √13. Irrational roots — factoring would have wasted time.
x = −3 ± √13Common Mistakes and How to Avoid Them
The quadratic formula is plug-and-chug, but small errors compound quickly. Here are the most common mistakes students make.
If b = −7, then −b = 7, not −7. Always wrap coefficients in parentheses when substituting.
The formula only works when the right side is 0. For 2x² + 3x = 5, rewrite as 2x² + 3x − 5 = 0 before identifying coefficients.
b² − 4ac(−7)² = 49, not −49. The square of a negative is positive.
The formula is x = (−b ± √(b² − 4ac))/(2a), with both the −b and the radical over 2a. Writing x = −b ± √(b² − 4ac)/(2a) divides only the radical part, which is wrong.
±A quadratic almost always has two roots (one when the discriminant is zero, complex pairs when negative). Computing only one solution loses half the answer.
√50 should be written as 5√2, not left as √50. Teachers expect simplified radical form.
Always treat the entire (−b ± √(b² − 4ac)) as a single numerator.
A simple guardrail: write the formula with explicit parentheses every time, and box your discriminant calculation separately.
Worked Examples and Real-World Applications
Solve 3x² − 5x + 1 = 0.
a = 3, b = −5, c = 1.
Discriminant: (−5)² − 4(3)(1) = 25 − 12 = 13.
Since 13 isn't a perfect square, the roots are irrational: x = (5 ± √13) / 6.
x ≈ 1.43 or x ≈ 0.23Solve x² + 4x + 8 = 0.
a = 1, b = 4, c = 8.
Discriminant: 16 − 32 = −16.
Two complex solutions: x = (−4 ± √(−16))/2 = (−4 ± 4i)/2 = −2 ± 2i.
x = −2 ± 2iA ball is thrown upward from a 6-foot platform with an initial velocity of 32 ft/s. Its height after t seconds is h(t) = −16t² + 32t + 6. When does it hit the ground?
Set h = 0: −16t² + 32t + 6 = 0, or equivalently 16t² − 32t − 6 = 0.
Using the formula with a = 16, b = −32, c = −6: discriminant = 1024 + 384 = 1408.
√1408 = 8√22 ≈ 37.5. t = (32 ± 37.5)/32.
The positive root is t ≈ 2.17 seconds. The negative root is rejected since time can't be negative.
t ≈ 2.17 secondsA company's revenue from selling x units is R(x) = −2x² + 100x. Find where revenue is zero.
Solving −2x² + 100x = 0 via the formula: a = −2, b = 100, c = 0.
Discriminant = 10000 − 0 = 10000. x = (−100 ± 100) / −4.
So x = 0 (no units sold) or x = 50 (price drops to zero).
x = 0 or x = 50A rectangular garden has area 60 sq ft and perimeter 32 ft. Find the dimensions.
Let length = ℓ and width = 16 − ℓ (from perimeter). Then ℓ(16 − ℓ) = 60, or ℓ² − 16ℓ + 60 = 0.
Discriminant = 256 − 240 = 16. ℓ = (16 ± 4)/2, so ℓ = 10 or ℓ = 6.
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