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AlgebraGrades 9–125 min read

Quadratic Formula

The quadratic formula solves every quadratic equation and reveals how its roots depend on the shape of its parabola.

Cheat sheet
The quadratic formula solves any equation of the form $ax^2+bx+c=0$, where $a\ne0$.

The idea in one minute

A quadratic equation asks where a parabola reaches height zero. Those locations are its roots, zeros, or solutions—three views of the same $x$-values.

Some quadratics factor neatly. Many do not. The quadratic formula works in both cases:

$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

First write the equation in standard form:

$$ ax^2+bx+c=0. $$

Then $a$, $b$, and $c$ are the coefficients, including their signs.

What each part means

SymbolMeaningIn $2x^2-7x-4=0$
$a$coefficient of $x^2$$2$
$b$coefficient of $x$$-7$
$c$constant term$-4$
$\pm$use a plus branch and a minus branchtwo possible roots
$b^2-4ac$the discriminant$81$

A visual reading

A parabola crossing the horizontal axis at x equals 1 and x equals 5, with its axis of symmetry halfway between the roots.

The two outputs are the two places where the parabola crosses the $x$-axis. Their midpoint is $-b/(2a)$, the $x$-coordinate of the vertex. The formula starts at that midpoint and moves an equal distance in the plus and minus directions.

A dependable five-step method

  1. Rearrange the equation so one side is zero.
  2. Record $a$, $b$, and $c$ with their signs.
  3. Calculate the discriminant $D=b^2-4ac$.
  4. Substitute into $x=(-b\pm\sqrt D)/(2a)$.
  5. Simplify both branches and check when practical.

Writing the coefficient list before substituting prevents most avoidable errors.

Worked example: two rational roots

Worked example: an irrational answer

The discriminant predicts the graph

Value of $D=b^2-4ac$Real rootsParabola
$D>0$two distinct rootscrosses the axis twice
$D=0$one repeated roottouches the axis once
$D<0$no real rootsdoes not meet the axis

When complex numbers are allowed, a negative discriminant gives two complex conjugate roots. “No real roots” is therefore more precise than “no solution.”

Why the formula works

Begin with $ax^2+bx+c=0$ and divide by $a$:

$$ x^2+\frac ba x=-\frac ca. $$

Complete the square by adding $(b/2a)^2$ to both sides:

$$ \left(x+\frac b{2a}\right)^2 =\frac{b^2-4ac}{4a^2}. $$

Take both square roots and isolate $x$:

$$ x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}, $$
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. $$

The quadratic formula is Completing the Square applied once to every possible quadratic.

Common mistakes

Choosing a solving method

  • Use factoring when the factors are visible.
  • Use square roots when the equation looks like $(x-h)^2=k$.
  • Use completing the square when vertex form matters.
  • Use the quadratic formula when you want a method that always works.

Applications

Quadratics model projectile height, maximum area or revenue, intersection points, and relationships with constant second differences. The two roots can represent two times, two positions, or two possible dimensions. Context decides whether both answers are meaningful.

Quick self-check

  • Is the equation equal to zero?
  • Did I keep every coefficient sign?
  • Did I calculate the discriminant separately?
  • Is the entire numerator over $2a$?
  • Did I use both branches?
  • Does substitution or the graph support the result?
Check your understanding

Try it yourself

Hints are part of learning. Open one whenever it makes the next step feel possible.

4 practice questions
Question 1Identify coefficients · Gentle

For x² − 6x + 5 = 0, which values should be used for a, b, and c?

Question 2Calculate a discriminant · Gentle

Find the discriminant of x² − 6x + 5 = 0.

Question 3Use the quadratic formula · Standard

Use the quadratic formula to find the larger root of x² − 6x + 5 = 0.

Question 4Interpret a discriminant · Standard

What does a discriminant of −12 tell you about the real roots?

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