What this guide helps you decide
A quadratic equation has the form ax² + bx + c = 0 with a nonzero leading coefficient. The discriminant b² − 4ac determines whether the roots are two distinct real values, one repeated real value, or a complex-conjugate pair.
The ± symbol creates two branches. Both must be evaluated and checked against the original equation; dropping a sign is a common source of incomplete answers.
Use the discriminant before evaluating roots
Calculate D = b² − 4ac, then x = (−b ± √D)/(2a). Keeping the numerator grouped avoids dividing only the radical by 2a.
- Move all terms to one side and identify a, b, and c with their signs.
- Reject a = 0 as a different equation class.
- Calculate the discriminant and choose the correct real or complex path.
- Evaluate both branches and substitute each result into the original equation.
Worked scenario: two integer roots
Solve x² − 7x + 10 = 0.
- a = 1, b = −7, c = 10; D = (−7)² − 4×1×10 = 9.
- x = (7 ± 3)/2, giving x = 5 and x = 2.
- Checks: 25 − 35 + 10 = 0 and 4 − 14 + 10 = 0.
Outcome: Both 2 and 5 satisfy the original equation. The discriminant of 9 correctly predicted two distinct real roots.
Quadratic-solution checklist
- Put the equation in standard form.
- Carry the signs of b and c accurately.
- Evaluate both ± branches.
- Keep the whole numerator over 2a.
- Substitute every reported root back into the original equation.
Limits and responsible use
- The dedicated calculator solves the quadratic form; it does not transform arbitrary symbolic statements or explain every algebraic rearrangement like a general CAS.
- Near-degenerate coefficients can be numerically sensitive, so very large or tiny values require careful precision and scaling.
Authoritative references
These links support the definitions, conventions, or safety boundaries used in this guide. CalculatorToolset wrote the explanation and example independently.
- College Algebra 2eOpenStax, Rice University
- Prealgebra 2eOpenStax, Rice University
Frequently asked questions
What happens when the discriminant is zero?
Both branches coincide, producing one repeated real root at −b/(2a).
Why must a be nonzero?
If a is zero, the x² term disappears and the equation is linear or lower order rather than quadratic.