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How to Solve Quadratic Equations

Solve quadratic equations step by step — the quadratic formula, factoring, and completing the square with worked examples, the discriminant, and a free quadratic formula calculator.

Updated 2026-06-26

Free calculators used in this guide

Quadratic Formula Calculator

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The squared term is what makes it quadratic — removing it would collapse the equation into a simple linear one. Quadratics appear everywhere: the arc of a cricket ball, the shape of a satellite dish, compound interest over two periods, and the dimensions of a room given its area. Knowing three reliable methods to solve them puts you in control of a surprisingly wide range of problems.

Step 1 — Write the Equation in Standard Form and Identify a, b, c

Before applying any method, rearrange the equation so that one side is zero. Take 2x² − 4x − 6 = 0 as the running example throughout this guide. Divide every term by 2 to keep numbers small:

x² − 2x − 3 = 0

Now read off the coefficients:

  • a = 1 (coefficient of x²)
  • b = −2 (coefficient of x)
  • c = −3 (constant term)

Keep the signs attached to the numbers — a sign error here will carry through every subsequent step.

Step 2 — Compute the Discriminant

Before choosing a method, calculate the discriminant: Δ = b² − 4ac.

For our example: Δ = (−2)² − 4(1)(−3) = 4 + 12 = 16

Discriminant Solutions
Δ > 0 Two distinct real solutions
Δ = 0 One real solution (double root)
Δ < 0 No real solutions (complex roots)

A positive discriminant of 16 tells us immediately that x² − 2x − 3 = 0 has two real solutions. The square root of 16 is a whole number, which also hints that factoring will work cleanly here.

Method 1 — Quadratic Formula (Always Works)

The quadratic formula is the universal method. It works for every quadratic equation, even when the roots are irrational or complex.

x = [−b ± √(b² − 4ac)] / 2a

Substituting a = 1, b = −2, c = −3:

x = [−(−2) ± √16] / 2(1) = [2 ± 4] / 2

  • x₁ = (2 + 4) / 2 = 3
  • x₂ = (2 − 4) / 2 = −1

For equations with large or decimal coefficients, skip the arithmetic and use the Quadratic Formula Calculator — enter a, b, and c and it returns both roots, the discriminant, and the vertex instantly.

Method 2 — Factoring (Fastest When It Works)

Factoring is the quickest route when the roots happen to be integers or simple fractions. The goal is to rewrite ax² + bx + c as a product of two linear factors, then apply the zero-product property.

For x² − 2x − 3 = 0, find two numbers that:

  • Multiply to c = −3
  • Add to b = −2

Those numbers are −3 and +1 (−3 × 1 = −3; −3 + 1 = −2).

Write the factored form: (x − 3)(x + 1) = 0

Set each factor to zero:

  • x − 3 = 0 → x = 3
  • x + 1 = 0 → x = −1

Same answers, less arithmetic. If you spend more than thirty seconds hunting for the pair, stop and use the formula instead — not every quadratic factors over the integers.

Method 3 — Completing the Square

Completing the square transforms the equation into a perfect square on the left side, making the square root step trivial. It is the algebraic proof behind the quadratic formula and is most useful when you also need the vertex of the parabola.

Starting from x² − 2x − 3 = 0:

  1. Move the constant to the right: x² − 2x = 3
  2. Add (b/2)² = (−2/2)² = 1 to both sides: x² − 2x + 1 = 4
  3. The left side is now a perfect square: (x − 1)² = 4
  4. Take the square root of both sides: x − 1 = ±2
  5. Solve: x = 1 + 2 = 3 or x = 1 − 2 = −1

The vertex of the parabola is at (1, −4), which you can read directly from step 3: h = 1, and k = −4 (the negative of the right-hand side before taking the root, adjusted for a).

Real-World Example — Projectile Motion

A ball is thrown upward from a height of 48 feet with an initial velocity of 32 ft/s. Its height at time t seconds is:

h = −16t² + 32t + 48

When does it hit the ground? Set h = 0:

−16t² + 32t + 48 = 0

Divide by −16: t² − 2t − 3 = 0

This is exactly the same equation solved above. The roots are t = 3 and t = −1. Since negative time has no physical meaning, the ball hits the ground at t = 3 seconds.

For higher-degree polynomial equations in problems like this, the Polynomial Calculator can handle the expanded forms automatically.

Choosing the Right Method

Situation Recommended method
Coefficients are large or non-integer Quadratic formula
Roots look like integers and ac is small Try factoring first
You need the vertex or vertex form Completing the square
You want an instant verified answer Quadratic Formula Calculator

Summary

Every quadratic equation in standard form ax² + bx + c = 0 can be solved with three techniques: the quadratic formula (always reliable), factoring (fast for clean integers), and completing the square (also reveals the vertex). The discriminant b² − 4ac tells you before you start whether to expect two real roots, one double root, or complex roots. For the worked example x² − 2x − 3 = 0 all three methods give x = 3 and x = −1, which confirms they are equivalent — choose whichever fits the numbers in front of you.

Frequently Asked Questions

The quadratic formula is x = [-b ± √(b² - 4ac)] / 2a, where a, b, and c are coefficients from the standard form ax² + bx + c = 0. First identify a, b, and c from your equation, then substitute them into the formula. Compute the discriminant b² - 4ac under the square root, then evaluate both the + and − versions to get your two solutions. It works for every quadratic equation, no matter how messy the numbers look.
Use factoring first when the coefficients are small integers and a quick mental search can find two numbers that multiply to ac and add to b. If that search takes more than a few seconds, switch to the quadratic formula — it always works regardless of whether the roots are integers, fractions, or irrational numbers. Completing the square is mainly useful when you need the vertex form of the parabola, not just the roots.
The discriminant is the expression b² - 4ac inside the square root of the quadratic formula. When it is positive, the equation has two distinct real solutions. When it equals zero, there is exactly one real solution — called a double root. When it is negative, the equation has no real solutions; the roots are complex (involving the imaginary unit i).
Complex roots appear when the discriminant b² - 4ac is negative, because you would need the square root of a negative number. They come in conjugate pairs of the form p + qi and p - qi, where i = √(-1). In most applied problems — projectile motion, area, finance — a negative discriminant means the scenario described has no real-world solution, so you stop there.
Quadratic equations model any situation involving a squared relationship: the height of a thrown ball over time, the area of a rectangular space given a fixed perimeter, profit functions in business, and the shape of satellite dish reflectors. In physics, kinematic equations for constant acceleration are quadratic in time. Engineers and architects use them routinely when designing parabolic structures.
Completing the square is most useful when you want to rewrite the quadratic in vertex form y = a(x - h)² + k, which immediately reveals the vertex (h, k) of the parabola. It is also the algebraic foundation behind the quadratic formula itself — the formula is derived by completing the square on the general form. For pure root-finding, however, factoring or the formula is usually faster.
Yes — the [Quadratic Formula Calculator](/quadratic-formula-calculator/) on this site lets you enter a, b, and c and instantly returns both roots, the discriminant value, and the vertex coordinates. It is especially useful when coefficients are large or decimal-valued, where manual arithmetic is error-prone. For polynomial equations of higher degree, the [Polynomial Calculator](/polynomial-calculator/) handles those as well.
A quadratic is a degree-2 polynomial, and the Fundamental Theorem of Algebra states that a degree-n polynomial has exactly n roots (counting multiplicity) over the complex numbers. The ± in the quadratic formula is what produces the two values. Geometrically, the parabola y = ax² + bx + c can cross the x-axis at zero, one, or two points depending on the discriminant.
The graph of y = ax² + bx + c is a parabola — a U-shaped curve that opens upward when a > 0 and downward when a < 0. The solutions of ax² + bx + c = 0 are exactly the x-coordinates where the parabola crosses the x-axis. The vertex of the parabola sits at x = -b / 2a, which is the midpoint of the two roots whenever they exist.
A linear equation has the form ax + b = 0 and its graph is a straight line — it has exactly one solution. A quadratic equation has the form ax² + bx + c = 0 and its graph is a parabola — it can have zero, one, or two real solutions. The key difference is the x² term: it introduces a squared relationship that the straight-line model cannot capture.
Vertex form is y = a(x - h)² + k, where (h, k) is the vertex of the parabola. It is obtained from standard form by completing the square. The two forms are algebraically identical — expanding vertex form always returns standard form. Vertex form makes it easy to read off the maximum or minimum value of the quadratic and to sketch the graph quickly.
Start by defining a variable for the unknown quantity, then translate the problem into the form ax² + bx + c = 0. Identify a, b, and c, compute the discriminant to know how many solutions exist, and solve using your preferred method. Always check whether both roots make physical sense — a negative time or a negative length is usually discarded. Substituting the answer back into the original scenario is the best way to verify.

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