How to Solve Quadratic Equations Step by Step

Degree 2 is the key: the highest power of x is 2. Examples:

• x² − 5x + 6 = 0 → a = 1, b = −5, c = 6
• 2x² + 3x − 2 = 0 → a = 2, b = 3, c = −2
• x² + 4 = 0 → a = 1, b = 0, c = 4
• −x² + 2x = 0 → a = −1, b = 2, c = 0

If a = 0, the equation is linear (bx + c = 0), not quadratic. Expand and rearrange first when you start from (x−1)(x+4) = 0 or from vertex form.

The graph y = ax² + bx + c opens upward when a > 0 and downward when a < 0. Roots are x-intercepts of that graph. The free quadratic equation solver reads a, b, c and returns the full solution package so you can match paper work to a parabola preview when available.

Word problems that become quadratics often hide in area (length × width), projectile height, or product constraints. Translate carefully before labeling coefficients.

For ax² + bx + c = 0:

x = (−b ± √(b² − 4ac)) / (2a)

1. Write the equation as ax² + bx + c = 0 (all terms on one side).
2. Identify a, b, c with correct signs.
3. Compute the discriminant D = b² − 4ac.
4. Take √D when D ≥ 0 (real roots) or proceed with i when D < 0 if complex roots are allowed.
5. Build both solutions: one with +√D, one with −√D (or a repeated root when D = 0).
6. Simplify fractions and radicals.

Worked example

x² − 5x + 6 = 0 → a = 1, b = −5, c = 6.

D = 25 − 24 = 1. x = (5 ± 1)/2 → x = 3 or x = 2.

Check: (x−2)(x−3) = x² − 5x + 6. Squares under the radical use ordinary power rules — the power guide covers √ and squares when radicals get messy.

Run the same coefficients through the free quadratic equation solver and compare both roots and the discriminant line-by-line.

The discriminant is D = b² − 4ac.

• D > 0 — two distinct real roots; parabola crosses the x-axis twice
• D = 0 — one real root (repeated/double root); parabola touches at the vertex
• D < 0 — no real roots; two complex conjugate roots if you work over the complex numbers

Examples

x² + 2x + 1 = 0 → D = 4 − 4 = 0 → x = −1 (double).

x² + 1 = 0 → D = −4 → x = ±i (complex).

2x² − 4x − 6 = 0 → D = 16 + 48 = 64 → x = (4 ± 8)/4 → x = 3 or x = −1.

Sign of a does not change how you classify D; it only flips which way the parabola opens. When D is a perfect square, roots are rational (if a, b, c are integers with gcd conditions); otherwise real roots may involve irrationals.

The solver’s discriminant analysis should match this three-way split. For comparing |root| sizes after you solve, use the absolute value calculator.

The axis of symmetry is the vertical line:

x = −b / (2a)

The vertex is at that x with y = f(x) = ax² + bx + c:

Vertex = (−b/(2a), f(−b/(2a)))

Vertex form: y = a(x − h)² + k with vertex (h, k). Completing the square converts standard form into vertex form.

Example

y = x² − 4x + 1 → h = −(−4)/(2) = 2. y(2) = 4 − 8 + 1 = −3. Vertex (2, −3). Axis x = 2.

When D > 0, the two roots are symmetric about the axis: their midpoint is h = −b/(2a). That is the same midpoint idea as averaging two x-coordinates — see the midpoint guide.

The free quadratic equation solver should report vertex, axis, and roots together so you can check that the roots straddle h when both are real.

Maximum or minimum value of the quadratic is the vertex y-coordinate (max if a < 0, min if a > 0).

Three standard real methods:

• Factoring — fastest when integers work: x² − 5x + 6 = (x−2)(x−3)
• Completing the square — builds vertex form and derives the quadratic formula
• Quadratic formula — always works for any real a, b, c with a ≠ 0

Completing the square for x² + bx: add (b/2)². For ax² + bx + c, factor a out of the x terms first, complete the square inside, then adjust constants.

Example: x² + 6x + 5 = 0 → (x+3)² − 9 + 5 = 0 → (x+3)² = 4 → x+3 = ±2 → x = −1 or x = −5.

If factoring stalls after a minute, switch to the formula. Graphing calculators estimate roots; algebra class usually wants exact form. Powers and squares in completing-the-square steps are ordinary arithmetic — confirm awkward squares with the power calculator.

Match factored roots to formula roots in the quadratic equation solver before you trust a factorization on a test.

Sign and structure errors dominate:

• Wrong sign on b when the formula uses −b — write b carefully from the equation
• Forgetting 2a in the denominator — apply it to the entire (−b ± √D), not only one part
• D = b² − 4ac with a minus slip — compute 4ac as a product first
• Dividing only one term by 2a
• Treating √(u+v) as √u + √v
• Dropping the ± and reporting one root only when D > 0

Equations like (2x−1)² = 9 need both 2x−1 = 3 and 2x−1 = −3. Absolute-value equations |x| = k are a related “both sides” pattern — the absolute value guide covers that structure.

Linear slope between two points on a parabola is not the parabola’s “slope” as a single m; local steepness changes. For straight-line slope only, use the slope calculator guide.

After solving, substitute each root back into ax² + bx + c and confirm you get ~0 (floating-point noise is tiny for clean integers).

Enter a, b, and c; read roots and structure.

1. Open the free quadratic equation solver.
2. Rearrange your equation to ax² + bx + c = 0.
3. Enter coefficients a, b, c (include negatives).
4. Read the discriminant and root classification.
5. Note real or complex roots and simplified forms when shown.
6. Review vertex form, axis of symmetry, and vertex coordinates.
7. Follow the step-by-step panel and compare with your paper work.
8. Substitute roots back into the original equation to verify.

Related tools: power calculator for squares and roots, absolute value calculator, slope calculator for lines, and midpoint calculator for root midpoints on the axis.