To solve quadratic equations, use the quadratic formula: x = (−b ± √(b² − 4ac)) / (2a). Plug in the coefficients from your equation, simplify, and you get the solutions — the values of x that make the equation equal zero. This formula works for every quadratic equation, no exceptions.
The quadratic formula is one of the most universally applicable tools in algebra. Unlike factoring — which only works cleanly when the equation has nice integer roots — the quadratic formula handles messy coefficients, irrational answers, and every case in between. It appears in physics (projectile motion), engineering (parabolic structures), and economics (profit maximization). Knowing how to apply it is a requirement for virtually every math course from algebra onward.
Use the free Quadratic Formula Calculator at RoughTools to solve any quadratic instantly with full step-by-step output — or follow the method below.
The Quadratic Formula
Every quadratic equation can be written in standard form: ax² + bx + c = 0, where a ≠ 0. Once your equation is in that form, you read off three coefficients and apply the formula.
Quadratic formula:
x = (−b ± √(b² − 4ac)) / (2a)
Where:
- a — the coefficient of x² (the squared term); must not be zero
- b — the coefficient of x (the linear term); can be zero or negative
- c — the constant term (no x attached); can be zero or negative
- ± — means you calculate two answers: one with addition (+) and one with subtraction (−)
- b² − 4ac — the discriminant, which determines how many real solutions exist
- √ — the square root of the discriminant
Worked example: 3x² − 7x − 5 = 0
Step 1 — Identify a, b, and c:
a = 3
b = −7
c = −5
Step 2 — Calculate the discriminant (b² − 4ac):
b² = (−7)² = 49
4ac = 4 × 3 × (−5) = −60
b² − 4ac = 49 − (−60) = 49 + 60 = 109
Step 3 — Take the square root of the discriminant:
√109 ≈ 10.440
Step 4 — Apply the quadratic formula (both solutions):
x = (−(−7) ± 10.440) / (2 × 3)
x = (7 ± 10.440) / 6
x₁ = (7 + 10.440) / 6 = 17.440 / 6 ≈ 2.907
x₂ = (7 − 10.440) / 6 = −3.440 / 6 ≈ −0.573
The result: the two solutions are x ≈ 2.907 and x ≈ −0.573. These are the two x-values where the parabola y = 3x² − 7x − 5 crosses the x-axis. Both answers are valid and both should be reported. Substituting either value back into the original equation produces (approximately) zero, which confirms the answers.
How to Solve Quadratic Equations Step by Step
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Write the equation in standard form: ax² + bx + c = 0. Every term must be on one side with zero on the other. If your equation is written as 3x² = 7x + 5, subtract 7x and add 5 to both sides: 3x² − 7x − 5 = 0. This step is easy to skip and causes errors downstream — the formula only works on standard form.
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Identify a, b, and c with their correct signs. The sign is part of the coefficient. For 3x² − 7x − 5 = 0: a = 3, b = −7, c = −5. Writing b = 7 or c = 5 (positive) is the single most common error in applying the quadratic formula. Double-check the sign attached to each term before plugging in.
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Calculate the discriminant: b² − 4ac. This single value tells you what kind of solutions to expect before you finish the calculation. A positive discriminant means two real solutions. Zero means exactly one solution. A negative discriminant means no real solutions (only complex ones). For our example: 49 − (−60) = 109. Positive — two real solutions confirmed.
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Take the square root of the discriminant. If the discriminant is a perfect square (4, 9, 25, 49...), the square root is a whole number and the solutions will be rational. If it is not a perfect square (like 109), the solutions are irrational — leave the answer in exact form (√109) or round to the required decimal places. For most school problems, round to three decimal places unless told otherwise.
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Apply the ± to get both solutions. Calculate x with addition first (x₁), then calculate x with subtraction (x₂). These are two separate division steps. Students who do both in one calculation almost always make a sign error. Treat them as two independent calculations that happen to share the same numerator components.
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Verify by substituting both answers back into the original equation. Plug each solution into ax² + bx + c and confirm the result is zero (or very close to zero for rounded answers). For x ≈ 2.907: 3(2.907)² − 7(2.907) − 5 = 25.35 − 20.35 − 5 ≈ 0. Correct. This check catches sign errors and arithmetic mistakes before you submit your work.
Pro tip: Always calculate b² before worrying about the −4ac part. Students who try to compute b² − 4ac in a single step frequently apply the negative incorrectly when b is negative. Squaring a negative number always gives a positive result — (−7)² = 49, never −49.
What Is the Discriminant and What Does It Tell You?
The discriminant — the expression b² − 4ac inside the square root — tells you how many real solutions a quadratic equation has before you finish solving it.
| Discriminant value | Number of solutions | Type of solutions | |---|---|---| | Positive (> 0) | Two | Two distinct real numbers | | Zero (= 0) | One | One repeated real number | | Negative (< 0) | Zero | Two complex (imaginary) numbers |
For 3x² − 7x − 5 = 0, the discriminant is 109 (positive), so there are two distinct real solutions. This matches what the parabola does geometrically: it crosses the x-axis at two separate points.
A discriminant of zero produces a repeated root — the parabola just touches the x-axis at its vertex without crossing. For example, x² − 6x + 9 = 0 has discriminant 36 − 36 = 0, and the single solution is x = 3. The parabola y = x² − 6x + 9 touches the x-axis at exactly (3, 0).
A negative discriminant means the parabola does not cross the x-axis at all — it sits entirely above or below zero. The solutions involve √(negative number), which produces imaginary numbers in the form a ± bi. These are valid mathematical answers but not real numbers. Most high school courses stop here and say "no real solutions." In more advanced contexts, you would continue to the complex solution: x = (7 ± i√|discriminant|) / (2a).
Checking the discriminant first is the fastest way to know what you are dealing with — it takes five seconds and prevents wasted work on equations with no real solutions.
When Can You Use Factoring Instead of the Quadratic Formula?
Factoring is faster than the quadratic formula when the equation has small integer roots — but it only works cleanly for that subset of problems.
To factor x² + 5x + 6 = 0: find two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3. Factored form: (x + 2)(x + 3) = 0. Solutions: x = −2 and x = −3.
Factoring works quickly when:
- The leading coefficient (a) is 1
- The constant (c) has few factors to check
- The discriminant is a perfect square
For the equation 3x² − 7x − 5 = 0, checking which factor pairs of −15 (the product of a and c) sum to −7 is trial-and-error that quickly becomes tedious. The quadratic formula produces the answer directly.
The guideline used by most algebra teachers: try factoring first for simple equations where you can spot the factors in under 30 seconds. If you cannot see the factors immediately, use the quadratic formula. There is no prize for factoring — the formula always gets you to the same answer.
Completing the square is a third method — it derives the quadratic formula algebraically and is useful for understanding where the formula comes from, but it is slower to apply than the formula itself in most cases.
The quadratic formula calculator shows which method applies to your specific equation and provides the full solution either way.
What Happens When the Discriminant Is Negative?
When b² − 4ac is negative, the quadratic equation has no real number solutions. The two solutions are complex numbers — numbers involving i, where i = √(−1).
For the equation x² + 4x + 7 = 0:
a = 1, b = 4, c = 7
Discriminant = 4² − 4(1)(7) = 16 − 28 = −12
Since −12 < 0, there are no real solutions. The complex solutions are:
x = (−4 ± √(−12)) / 2
x = (−4 ± √12 × √(−1)) / 2
x = (−4 ± 2√3 × i) / 2
x = −2 ± i√3
The two complex solutions are x = −2 + i√3 and x = −2 − i√3. These always come in conjugate pairs — one with +i and one with −i — when the original coefficients are real numbers.
Geometrically, a negative discriminant means the parabola does not intersect the x-axis. If a is positive, the entire parabola sits above the x-axis (always positive values). If a is negative, it sits entirely below. In either case, there is no x-value where y = 0.
For most high school algebra courses, a negative discriminant ends the problem: you state "no real solutions." In pre-calculus and beyond, you continue to the complex form. Always report whether your answer is real or complex — leaving just "x = ±i√3" without context is incomplete.
Common Mistakes to Avoid When Solving Quadratic Equations
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Forgetting to write the equation in standard form first. The formula requires ax² + bx + c = 0, with zero on the right. An equation like 2x² + 3x = 8 must be rewritten as 2x² + 3x − 8 = 0 before identifying a, b, and c. Reading coefficients from a non-standard form produces wrong values and wrong answers.
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Squaring a negative b incorrectly. When b is negative, b² is positive — always. A common error: writing (−7)² = −49. This is wrong. (−7)² = (−7) × (−7) = 49. A negative number squared is never negative. If your discriminant comes out negative but b² is large, recheck this step.
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Dropping the negative sign from −b at the front of the formula. The formula starts with −b, not b. If b = −7, then −b = −(−7) = +7. Students who write "b = −7, so the formula starts with −7" make this error. The formula negates b — a negative b becomes positive at the front of the formula.
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Dividing only the numerator's first term by 2a instead of the whole numerator. The entire expression (−b ± √discriminant) is divided by 2a, not just the −b part. Writing x = −b/2a ± √discriminant is a structural error. Use parentheses to keep the full numerator grouped: (−b ± √discriminant) / (2a).
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Reporting only one solution when two exist. A positive discriminant always produces two distinct solutions. Stopping at x₁ and failing to compute x₂ means your answer is half-complete. Exams award separate marks for each solution — missing one costs points even if the other is correct.
Frequently Asked Questions
What does the ± symbol mean in the quadratic formula? The ± (plus-or-minus) symbol means you perform the calculation twice — once adding the square root and once subtracting it. This gives two solutions: x₁ = (−b + √discriminant) / (2a) and x₂ = (−b − √discriminant) / (2a). Both are valid answers. The ± exists because quadratic equations are degree-2 polynomials, and a degree-2 polynomial has at most two roots.
What if my quadratic equation is missing the b or c term? If b = 0, the equation is in the form ax² + c = 0. Substitute b = 0 into the formula: x = (0 ± √(0 − 4ac)) / (2a) = ± √(−4ac) / (2a). It is often faster to solve directly: ax² = −c → x² = −c/a → x = ± √(−c/a). If c = 0, the equation is ax² + bx = 0, which factors as x(ax + b) = 0, giving x = 0 and x = −b/a. The formula handles both cases correctly — substituting zero is always valid.
What is the difference between the quadratic formula and completing the square? Completing the square is an algebraic process that rewrites ax² + bx + c = 0 into a form where you can take the square root directly. The quadratic formula was actually derived by completing the square on the general form — it is the pre-solved result. In practice: use completing the square to understand the derivation or to convert a quadratic to vertex form (y = a(x − h)² + k). Use the quadratic formula when you just need the roots quickly.
How many solutions can a quadratic equation have? A quadratic equation can have zero, one, or two real solutions — determined entirely by the discriminant. Two solutions when b² − 4ac > 0. Exactly one when b² − 4ac = 0. Zero real solutions when b² − 4ac < 0 (though two complex solutions always exist). A quadratic cannot have three or more solutions — by the Fundamental Theorem of Algebra, a degree-2 polynomial has exactly 2 roots counting multiplicity, real or complex.
When should I use the quadratic formula vs the scientific calculator? Use the quadratic formula calculator when you need the answer quickly, want to verify your manual work, or are dealing with large or decimal coefficients where arithmetic errors are likely. Work through the formula by hand when you need to show your steps on an exam or homework, or when you want to understand why the answers are what they are. Many exams require shown work — the calculator gives the answer, but the marks come from the steps. Use the scientific calculator for the arithmetic within each step if needed.
Use the Free Quadratic Formula Calculator
The Free Quadratic Formula Calculator at RoughTools solves any quadratic equation in standard form — enter a, b, and c, and get both solutions instantly. It shows the full step-by-step working including the discriminant, square root calculation, and both final answers in exact and decimal form. It also flags when the discriminant is negative, showing complex solutions rather than returning an error. No account needed, no data stored, completely free.
Free Quadratic Formula Calculator →
You might also need:
- Scientific Calculator — evaluate square roots, exponents, and complex arithmetic within each formula step
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