Algebra⏱ 7 min read

Quadratic Formula

x = (-b ± √(b² - 4ac)) / 2a

What is the Quadratic Formula?

The quadratic formula solves any equation of the form ax² + bx + c = 0 for x, regardless of whether it factors neatly. It comes from completing the square on the general quadratic — a derivation worth doing once by hand, since it explains why the formula has the shape it does (see below).

The expression under the square root, b² − 4ac, is called the discriminant. Its sign tells you what kind of solutions to expect before you finish the calculation: positive means two distinct real roots, zero means one repeated real root, and negative means no real roots (the parabola never crosses the x-axis).

What Each Variable Means

Root(s) of the equationThe value(s) of x that satisfy ax² + bx + c = 0.

Quadratic coefficientThe coefficient of x². Cannot be zero — otherwise the equation isn't quadratic.

Linear coefficientThe coefficient of x, the equation's middle term.

Constant termThe term with no x attached.

When to Use It

Solving any equation in the form ax² + bx + c = 0
When factoring by inspection isn't obvious or possible
Checking whether a quadratic has real or complex roots before graphing it

Where This Formula Comes From

Start with the general quadratic

Every quadratic equation can be written in this standard form.

ax² + bx + c = 0

Divide every term by a

This makes the x² coefficient 1, which is required to complete the square.

x² + (b/a)x + c/a = 0

Move the constant to the right side

Isolate the x terms.

x² + (b/a)x = -c/a

Complete the square

Add (b/2a)² to both sides — the amount needed to make the left side a perfect square trinomial.

x² + (b/a)x + (b/2a)² = (b/2a)² - c/a

Factor the left side

The left side is now a perfect square; simplify the right side over a common denominator.

(x + b/2a)² = (b² - 4ac) / 4a²

Take the square root of both sides

Remember both the positive and negative root.

x + b/2a = ± √(b² - 4ac) / 2a

Isolate x

Subtract b/2a from both sides to get the final formula.

x = (-b ± √(b² - 4ac)) / 2a

Step-by-Step Examples

Example 1: Two real roots

Problem: x² - 5x + 6 = 0

Identify a, b, and c

Match the equation to the general form ax² + bx + c = 0.

a = 1, b = -5, c = 6

Calculate the discriminant

b² - 4ac tells you how many real solutions exist.

(-5)² - 4(1)(6) = 25 - 24 = 1

Substitute into the formula

Plug a, b, and the discriminant into x = (-b ± √(b²-4ac)) / 2a.

x = (5 ± √1) / 2 = (5 ± 1) / 2

Compute both roots

Solve once with + and once with −.

x₁ = (5 + 1) / 2 = 3 x₂ = (5 - 1) / 2 = 2

✓

Answer: x = 3 or x = 2 — verified: (3)² - 5(3) + 6 = 0 and (2)² - 5(2) + 6 = 0

Example 2: No real roots

Problem: 2x² + 4x + 5 = 0

Identify a, b, and c

Match the equation to the general form.

a = 2, b = 4, c = 5

Calculate the discriminant

Check its sign before going further.

4² - 4(2)(5) = 16 - 40 = -24

✓

Answer: No real solutions — the discriminant is negative, so the square root has no real value. (The two complex roots are x = -1 ± i√3/√2, if complex numbers are in scope.)

Interactive Calculator

a (x² coefficient)b (x coefficient)c (constant)

Result will appear here

Common Mistakes

Mistake: Forgetting the ± and reporting only one root.
Fix: A positive discriminant always gives two roots — compute both (-b+√disc)/2a and (-b-√disc)/2a.
Mistake: Dividing only the square-root term by 2a instead of the whole numerator.
Fix: The entire numerator, -b ± √(b²-4ac), is divided by 2a — not just the radical.
Mistake: Losing the sign of b when substituting.
Fix: If the equation is x² - 5x + 6 = 0, then b = -5. Substitute the coefficient's actual sign, not its magnitude.

Practice Questions

Solve x² - 3x - 10 = 0 using the quadratic formula.
Hint: a = 1, b = -3, c = -10.
How many real solutions does 2x² + 4x + 5 = 0 have?
Hint: Compute the discriminant before trying to solve.

Frequently Asked Questions

What if a = 0?

Then it isn't a quadratic equation — it's linear (bx + c = 0), and the quadratic formula doesn't apply. Solve linear equations directly instead.

Does the quadratic formula work for every quadratic equation?

Yes — unlike factoring, which only works cleanly for equations with rational roots, the quadratic formula solves any quadratic, including ones with irrational or complex roots.

Why does a negative discriminant mean no real solutions?

Because the formula requires taking the square root of the discriminant, and no real number squares to a negative value. The parabola simply never crosses the x-axis.