Quadratic Equation Solver

Worked examples

What the discriminant tells you

Textbook formula against the stable formula

How to solve a quadratic equation

1. Write it as ax² + bx + c = 0. Move every term to one side so the other side is 0: 2x^2 - 4x = 6 becomes 2x² − 4x − 6 = 0, so a = 2, b = −4 and c = −6.
2. Find the discriminant. Compute D = b² − 4ac. For a = 2, b = −4, c = −6: D = b² − 4ac = (−4)² − 4 × 2 × (−6) = 16 + 48 = 64.
3. Read the sign of D. If D > 0 there are two real roots, if D = 0 one repeated root, and if D < 0 two complex roots. Here D = 64, which is a perfect square, so the roots are rational.
4. Apply the quadratic formula. x = (−b ± √D) / (2a) = (4 ± √64) / 4 = (4 ± 8) / 4, so x = −1 or 3. Simplify any radical and cancel: (−6 ± √20) / 4 = (−6 ± 2√5) / 4 = (−3 ± √5) / 2.
5. Check with Vieta’s formulas. The roots add up to −b/a = 2 and multiply to c/a = −3. For x = −1 or 3: −1 + 3 = 2 and −1 × 3 = −3.
6. Find the vertex and graph it. The vertex is at x = −b / (2a) = 1, and its height is f(x) there, giving (1, −8). Plot the vertex, the roots and the y-intercept (0, −6).

Frequently asked questions

How do you solve a quadratic equation?

Write it as ax² + bx + c = 0, find the discriminant D = b² − 4ac, and use x = (−b ± √D) / (2a). For 2x² − 4x − 6 = 0, D = 64, so x = (4 ± 8) / 4, which is −1 or 3. If the quadratic factors you can also factor it, here as 2(x + 1)(x − 3), and set each factor to 0.

What is the discriminant, and what does it tell you?

The discriminant is D = b² − 4ac, the part under the square root in the quadratic formula. D > 0 gives two real roots (rational when D is a perfect square, like 64 for 2x² − 4x − 6 = 0), D = 0 gives one repeated root (4x² − 12x + 9 = 0 has x = 3/2), and D < 0 gives two complex conjugate roots (x² + 2x + 5 = 0 has D = −16).

How do you find complex roots?

When D is negative, √D is an imaginary number: √(−16) = 4i. The roots are then −b/(2a) ± (√(−D)/(2a))·i. For x² + 2x + 5 = 0, x = (−2 ± √(−16)) / 2 = (−2 ± 4i) / 2 = −1 ± 2i. The two roots are complex conjugates, and x² + 4 = 0 gives the pure imaginary pair ±2i.

How do I get an exact answer like (−3 ± 2√5)/4?

The tool keeps the discriminant as an exact whole number, pulls out square factors from the radical (√1280 = 16√5), and divides the top and bottom by their common factor. 16x² + 24x − 11 = 0 has D = 1280 and x = (−3 ± 2√5)/4 (x₁ ≈ −1.868033989, x₂ ≈ 0.3680339887). Fractions and decimals in a, b and c are fine: they are cleared with a common multiplier first, so (1/2)x² − (3/2)x + 1 = 0 gives x = 1 or x = 2 and nothing is rounded.

What if a is 0?

Then there is no x² term and the equation is linear, so the quadratic formula (which divides by 2a) does not apply. The solver switches to bx + c = 0 and gives x = −c/b: 2x − 3 = 0 gives x = 3/2. If a and b are both 0 the equation is either 0 = 0, true for every x, or a false statement such as 5 = 0, which has no solution.

How do I find the vertex and the axis of symmetry?

The vertex is at x = −b / (2a), halfway between the roots, and its height is the value of the quadratic there. For 2x² − 4x − 6 = 0 that is (1, −8). The axis of symmetry is the vertical line through it, x = 1. The vertex is the minimum when a > 0 (the parabola opens upward) and the maximum when a < 0. In vertex form the equation is y = 2(x − 1)² − 8.

How does completing the square work?

Divide by a, move the constant to the right, and add (b/2a)² to both sides so the left side is a perfect square: 2x² − 4x − 6 = 0 becomes (x − 1)² = 4, so x − 1 = ±2 and x = 1 ± 2 = −1 or 3. Doing this with letters instead of numbers gives the quadratic formula. The tool writes out the steps for your own coefficients.

What are Vieta’s formulas?

For the roots x₁ and x₂ of ax² + bx + c = 0, x₁ + x₂ = −b/a and x₁ × x₂ = c/a. For 2x² − 4x − 6 = 0 the sum is 2 and the product is −3, matching −1 + 3 and −1 × 3. They let you check roots, or build a quadratic from two numbers: the sum 5 and product 6 give x² − 5x + 6 = 0, with roots 2 and 3.

Why do the decimals use a different formula from the one in the textbook?

When b² is much larger than |4ac|, one of (−b ± √D) is a subtraction of nearly equal numbers, and a computer loses most of its digits (catastrophic cancellation). For x² − 100,000,000x + 1 = 0 the textbook formula in double precision gives 7.45058×10⁻⁹ for the smaller root, off by a relative 0.25, while the stable form q = −½(b + sgn(b)√D), x₁ = q/a, x₂ = c/q gives 1×10⁻⁸. The tool computes the discriminant exactly first and uses the stable form for decimals (Press et al., Numerical Recipes, section 5.6; Higham, Accuracy and Stability of Numerical Algorithms, section 1.8).

How do I type an equation instead of coefficients?

Choose “An equation” and type it as you would write it, such as 2x^2 - 4x = 6 (the tool moves it to 2x² − 4x − 6 = 0). Terms may be on both sides: 3x^2 + 2x = x^2 - 5 becomes 2x² + 2x + 5 = 0. Use ^ or ² for the square, and parentheses around a fraction in front of x, like (3/4)x. Powers above 2 are refused because the equation would not be a quadratic.

Can every quadratic be factored?

Over the rational numbers only when the discriminant is a perfect square (the roots are then rational). 2x² − 4x − 6 = 0 factors as 2(x + 1)(x − 3); 2x² + 3x − 4 = 0 does not factor with rational numbers because its roots involve √41. 4x² − 12x + 9 = 0 is a perfect square: 4(x − 3/2)².

Are the answers exact, or rounded?

The discriminant, the exact roots, the vertex, the factored form and Vieta’s sums are exact: every coefficient is kept as a fraction of two whole numbers (up to 30 digits each). Only the lines marked ≈ are decimals, shown to 10 significant digits. A radical so large that its square factors cannot all be found is left as it stands, which is still correct but may not be fully simplified.