Quadratic Formula Calculator

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Algebra & Trig#28

Quadratic Formula Solver

Solves \(ax^2 + bx + c = 0\): discriminant, roots (real or complex), vertex and factored form.

This quadratic formula calculator solves any equation of the form \( ax^2 + bx + c = 0 \). Enter the three coefficients and it returns the discriminant, both roots in exact form and as decimals, the factored form, the vertex, the axis of symmetry and the y-intercept. When the discriminant is negative it gives the complex roots \( p \pm qi \) instead of an error.

It is built for algebra, pre-calculus and calculus students checking homework, and for anyone who needs the zeros of a parabola quickly. It’s free and needs no sign-up.

How to use the quadratic formula calculator

  1. Rearrange your equation so one side is zero: \( ax^2 + bx + c = 0 \).
  2. Enter a, b and c. Include the signs: for \( 2x^2 - 7x + 3 = 0 \), enter \( a = 2 \), \( b = -7 \), \( c = 3 \). Fractions such as 1/2 and decimals are accepted.
  3. Press Solve. The first row is the discriminant \( \Delta = b^2 - 4ac \), which tells you how many real roots there are.
  4. Read the roots, the factored form (when the roots are rational) and the vertex. The vertex is the minimum point when \( a > 0 \) and the maximum point when \( a < 0 \).

Formula

The calculator uses the quadratic formula, which comes from completing the square on \( ax^2 + bx + c = 0 \):

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

The expression under the root is the discriminant:

DiscriminantRootsGraph
\( \Delta > 0 \)Two different real rootsCrosses the x-axis twice
\( \Delta = 0 \)One repeated real rootTouches the x-axis at the vertex
\( \Delta < 0 \)Two complex rootsNever touches the x-axis

The vertex sits halfway between the roots, at

$$x_v = -\frac{b}{2a}, \qquad y_v = c - \frac{b^2}{4a}$$

Worked example: two rational roots

Solve \( 2x^2 - 7x + 3 = 0 \). These are the values loaded in the calculator above.

Discriminant. \( \Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25 \). It’s a perfect square, so the roots are rational.

Roots. \( x = \frac{7 \pm 5}{4} \), so \( x_1 = \frac12 \) and \( x_2 = 3 \).

Factored form. \( 2\left(x - \tfrac12\right)(x - 3) = (2x - 1)(x - 3) \).

Vertex. \( x_v = \frac{7}{4} \) and \( y_v = 3 - \frac{49}{8} = -\frac{25}{8} \). The parabola opens upward, so \( -\frac{25}{8} \) is its minimum value.

Worked example: complex roots

Solve \( x^2 + 2x + 5 = 0 \). The discriminant is \( 4 - 20 = -16 \), so there are no real roots. The formula still works with \( \sqrt{-16} = 4i \):

$$x = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

The vertex is \( (-1, 4) \). It lies above the x-axis and the parabola opens upward, which is why the graph never crosses it.

Worked example: irrational roots

For \( 2x^2 + 4x - 1 = 0 \), \( \Delta = 16 + 8 = 24 \). The calculator shows the exact form \( x = \frac{-4 \pm 2\sqrt6}{4} \), which simplifies to \( \frac{-2 \pm \sqrt6}{2} \), and the decimals \( x \approx 0.2247 \) and \( x \approx -2.2247 \).

Tips and common mistakes

  • Dropping a sign. In \( 3x^2 - 5x - 2 = 0 \), \( b = -5 \), so \( -b = 5 \) and \( b^2 = 25 \), not \( -25 \).
  • Dividing only part of the top by 2a. The whole numerator \( -b \pm \sqrt{\Delta} \) is divided by \( 2a \).
  • Forgetting to set the equation to zero. \( x^2 + 3x = 10 \) must become \( x^2 + 3x - 10 = 0 \) before you read off \( c = -10 \).
  • Using the formula when a = 0. Then the equation is linear, \( bx + c = 0 \), with the single root \( x = -c/b \).
  • Check by substitution. Plugging \( x = 3 \) into \( 2x^2 - 7x + 3 \) gives \( 18 - 21 + 3 = 0 \).

Where quadratics show up in calculus

Setting a derivative equal to zero often produces a quadratic. For \( f(x) = 2x^3 - 3x^2 - 12x \), \( f'(x) = 6x^2 - 6x - 12 = 0 \) gives \( x = -1 \) and \( x = 2 \), the critical points (see how to find critical points). Projectile heights \( h(t) = -4.9t^2 + v_0t + h_0 \) are quadratics too, and the projectile motion calculator solves them for you.

Related guides

Further reading

FAQ

What is the quadratic formula?

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It gives both solutions of \( ax^2 + bx + c = 0 \) for any \( a \neq 0 \).

How do I know how many solutions a quadratic has?

Look at the discriminant \( b^2 - 4ac \). Positive means two real solutions, zero means one repeated solution, and negative means two complex solutions.

Can the calculator give complex roots?

Yes. When \( b^2 - 4ac < 0 \), it shows the roots as \( p \pm qi \), for example \( -1 \pm 2i \) for \( x^2 + 2x + 5 = 0 \).

How do I find the vertex of a parabola?

Use \( x = -\frac{b}{2a} \), then substitute that x into the equation to get y. The calculator does both and also states whether the vertex is a minimum or a maximum.

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