What does the discriminant tell you?

Δ > 0: two distinct real roots (parabola crosses x-axis twice). Δ = 0: one repeated real root (parabola touches x-axis at the vertex). Δ < 0: two complex (imaginary) roots (parabola never crosses x-axis).

When should I factor instead of using the formula?

Factoring is faster when integers work out cleanly (like x²−5x+6=(x−2)(x−3)). If the coefficients are decimals, fractions, or don't factor neatly, go straight to the quadratic formula.

Can I solve cubic or higher-degree equations with this?

No. The quadratic formula works only for degree-2 equations. Cubic equations (x³) have a cubic formula, but degree 5 and above have no general algebraic solution — numerical methods are used instead.

What are complex roots used for in practice?

Complex roots appear in AC circuit analysis (impedance), signal processing (filter design), and quantum mechanics. In those contexts, complex numbers are not abstract — they represent physical oscillating quantities.

How do I convert a quadratic from vertex form to standard form?

Vertex form: a(x−h)²+k. Expand: a(x²−2hx+h²)+k = ax²−2ahx+ah²+k. So b=−2ah and c=ah²+k. Plug into the quadratic formula or use the vertex directly: roots are h ± √(−k/a).

What is the sum and product of the roots?

By Vieta's formulas: sum of roots = −b/a, product of roots = c/a. These let you quickly verify your answers: for x²−5x+6=0, sum should be 5 (2+3=5 ✓), product should be 6 (2×3=6 ✓).

Quadratic Equation Solver

Comprehensive Calculator Guide

The Quadratic Equation Solver finds the roots of any equation in the form ax² + bx + c = 0 using the quadratic formula. It shows the discriminant value, determines whether roots are real or complex, and displays full step-by-step working — ideal for students checking homework, verifying exam answers, or understanding the method.

Understanding the Quadratic Formula and Discriminant

Any equation of the form ax² + bx + c = 0 (where a ≠ 0) can be solved with the quadratic formula: x = (−b ± √(b²−4ac)) / 2a. The key to the formula is the discriminant Δ = b²−4ac under the square root. If Δ > 0, the equation has two distinct real roots. If Δ = 0, it has exactly one real root (a repeated root). If Δ < 0, the roots are complex conjugates — they involve the imaginary unit i = √(−1) — and the parabola does not cross the x-axis.

Why does the ± appear? Because a parabola (the graph of ax²+bx+c) is symmetric about its vertex. The two roots are equidistant from the axis of symmetry at x = −b/2a. The ± captures both roots in one formula. Factoring is faster when it works (x²−5x+6 = (x−2)(x−3) is obvious), but the quadratic formula always works for any valid coefficients.

Real-World Applications of Quadratic Equations

Physics: projectile motion is governed by quadratics. If a ball is thrown upward at 20 m/s from a 5 m ledge, its height at time t is h(t) = −4.9t²+20t+5. Setting h=0 and solving gives when it hits the ground. Engineering: structural load calculations, lens optics (focal length equations), and circuit resonance frequencies all produce quadratic equations.

Business and economics: profit maximization problems often involve quadratics. If profit P = −2x²+80x−300 where x is units sold, setting P=0 gives breakeven points and the vertex gives the maximum profit quantity. Geometry: finding the dimensions of a rectangle given its area and perimeter leads directly to a quadratic. The solver handles all these scenarios with any real or complex coefficients.

How to Use

01Enter coefficient a (must be non-zero)
02Enter coefficient b
03Enter coefficient c
04Click Solve to see roots, discriminant, and step-by-step working

Formula Used

x = (−b ± √(b²−4ac)) / 2a. Discriminant Δ = b²−4ac. Vertex at x = −b/2a.

Practical Examples

Example 1: Two real roots — x² − 5x + 6 = 0

a=1, b=−5, c=6. Δ=25−24=1. x=(5±1)/2. Roots: x=3 and x=2.

Example 2: Repeated root — x² − 6x + 9 = 0

a=1, b=−6, c=9. Δ=36−36=0. x=6/2=3. One root: x=3 (double root, parabola touches x-axis at one point).

Example 3: Complex roots — x² + 2x + 5 = 0

a=1, b=2, c=5. Δ=4−20=−16. x=(−2±√(−16))/2=(−2±4i)/2. Roots: x=−1+2i and x=−1−2i.

Important Tips

Before using the formula, check if the equation is easily factorable — it's faster. Look for two numbers that multiply to c and add to b (when a=1).
If you get complex roots, they always appear as conjugate pairs (a+bi and a−bi) — the graph of the parabola simply doesn't cross the x-axis.
Always verify roots by substituting back into the original equation: if ax²+bx+c = 0, you should get exactly zero.

Common Mistakes to Avoid

Forgetting the ±: the formula produces two roots. Writing x = (−b + √Δ)/2a only gives you one of them.
Using a=0: if the x² coefficient is zero, the equation is linear (bx+c=0), not quadratic. The formula is undefined for a=0.

Frequently Asked Questions

What does the discriminant tell you?: Δ > 0: two distinct real roots (parabola crosses x-axis twice). Δ = 0: one repeated real root (parabola touches x-axis at the vertex). Δ < 0: two complex (imaginary) roots (parabola never crosses x-axis).
When should I factor instead of using the formula?: Factoring is faster when integers work out cleanly (like x²−5x+6=(x−2)(x−3)). If the coefficients are decimals, fractions, or don't factor neatly, go straight to the quadratic formula.
Can I solve cubic or higher-degree equations with this?: No. The quadratic formula works only for degree-2 equations. Cubic equations (x³) have a cubic formula, but degree 5 and above have no general algebraic solution — numerical methods are used instead.
What are complex roots used for in practice?: Complex roots appear in AC circuit analysis (impedance), signal processing (filter design), and quantum mechanics. In those contexts, complex numbers are not abstract — they represent physical oscillating quantities.
How do I convert a quadratic from vertex form to standard form?: Vertex form: a(x−h)²+k. Expand: a(x²−2hx+h²)+k = ax²−2ahx+ah²+k. So b=−2ah and c=ah²+k. Plug into the quadratic formula or use the vertex directly: roots are h ± √(−k/a).
What is the sum and product of the roots?: By Vieta's formulas: sum of roots = −b/a, product of roots = c/a. These let you quickly verify your answers: for x²−5x+6=0, sum should be 5 (2+3=5 ✓), product should be 6 (2×3=6 ✓).