Algebra 1 · Algebra 2 · Grades 8, 9, 10

The Quadratic Formula: How and Why It Works

Quick answer

The quadratic formula solves any equation of the form ax² + bx + c = 0 with a ≠ 0. The solutions are x = (−b ± √(b² − 4ac)) / (2a). The expression under the radical, b² − 4ac, is the discriminant: if it is positive there are two real solutions, if it is zero there is one, and if it is negative there are no real solutions.

What you'll learn

  • Use the quadratic formula to solve any quadratic equation
  • Explain where the quadratic formula comes from
  • Use the discriminant to predict how many real solutions an equation has

What the quadratic formula is

A quadratic equation is any equation you can write in the form

ax2+bx+c=0,a≠0ax^2 + bx + c = 0, \qquad a \neq 0

The quadratic formula gives you both solutions directly from aa, bb, and cc:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

That ±\pm is doing real work. It is shorthand for two separate solutions:

x1=−b+b2−4ac2aandx2=−b−b2−4ac2ax_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \qquad\text{and}\qquad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

Why it works

Most courses hand you the formula and move on. It is worth seeing where it comes from, because the derivation is completing the square done once with letters instead of numbers. Doing it once with letters is what saves you from doing it every time with numbers.

Start with the general equation and divide through by aa (allowed, since a≠0a \neq 0):

x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Move the constant across:

x2+bax=−cax^2 + \frac{b}{a}x = -\frac{c}{a}

Now complete the square. Take half of the xx-coefficient, b2a\tfrac{b}{2a}, square it, and add it to both sides:

x2+bax+(b2a)2=−ca+(b2a)2x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^{2} = -\frac{c}{a} + \left(\frac{b}{2a}\right)^{2}

The left side is now a perfect square, and the right side combines over a common denominator:

(x+b2a)2=b2−4ac4a2\left(x + \frac{b}{2a}\right)^{2} = \frac{b^2 - 4ac}{4a^2}

Take the square root of both sides — and this is exactly where the ±\pm enters, because a positive number has two square roots:

x+b2a=± b2−4ac2ax + \frac{b}{2a} = \pm\,\frac{\sqrt{b^2 - 4ac}}{2a}

Finally, isolate xx:

x=−b2a±b2−4ac2a=−b±b2−4ac2ax = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

That is the whole derivation. Nothing in it is a trick you have not already seen.

How to use it

  1. Rearrange the equation so that one side is 00.
  2. Identify aa, bb, and cc, keeping their signs.
  3. Compute the discriminant b2−4acb^2 - 4ac first, on its own.
  4. Substitute into the formula and simplify.
  5. Check at least one solution by substituting it back in.

Worked examples

The discriminant

The quantity Δ=b2−4ac\Delta = b^2 - 4ac tells you how many real solutions exist before you finish solving:

Δ=b2−4ac{>0two distinct real solutions=0exactly one real solution<0no real solutions\Delta = b^2 - 4ac \quad \begin{cases} > 0 & \text{two distinct real solutions} \\[2pt] = 0 & \text{exactly one real solution} \\[2pt] < 0 & \text{no real solutions} \end{cases}

Practice problems

  1. Solve x2+7x+12=0x^2 + 7x + 12 = 0.

    Hint

    Compute the discriminant first. It is a perfect square, so the solutions are rational.

    Answer

    x=−3x = -3 or x=−4x = -4

    Full solution

    a=1a = 1, b=7b = 7, c=12c = 12.

    b2−4ac=49−48=1b^2 - 4ac = 49 - 48 = 1

    x=−7±12x = \tfrac{-7 \pm 1}{2}, giving x=−3x = -3 or x=−4x = -4.

    Check: (−3)2+7(−3)+12=9−21+12=0(-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0 ✓

  2. Solve 3x2−2x−5=03x^2 - 2x - 5 = 0.

    Answer

    x=53x = \tfrac{5}{3} or x=−1x = -1

    Full solution

    a=3a = 3, b=−2b = -2, c=−5c = -5.

    b2−4ac=4+60=64b^2 - 4ac = 4 + 60 = 64

    x=2±86x = \tfrac{2 \pm 8}{6}, giving x=106=53x = \tfrac{10}{6} = \tfrac{5}{3} or x=−66=−1x = \tfrac{-6}{6} = -1.

  3. How many real solutions does x2+2x+5=0x^2 + 2x + 5 = 0 have?

    Answer

    None.

    Full solution

    b2−4ac=4−20=−16b^2 - 4ac = 4 - 20 = -16, which is negative, so there are no real solutions. The parabola sits entirely above the xx-axis.

Frequently asked questions

When should I use the quadratic formula instead of factoring?

Try factoring first if you can spot two whole numbers that work. Use the quadratic formula when you cannot find that pair, when the solutions are irrational, or when you want a method that always works. The formula never fails; factoring only works for some equations.

What does it mean when the discriminant is negative?

A negative discriminant means the parabola never crosses the x-axis, so the equation has no real solutions. It does have two complex solutions, which you study in Algebra 2.

Key terms in this lesson

Discriminant
The discriminant is the expression b squared minus 4ac inside the quadratic formula. Its sign tells you how many real solutions a quadratic has before you finish solving.
Quadratic equation
A quadratic equation is one that can be written as ax squared plus bx plus c = 0, with a not equal to zero. The squared term is what makes it quadratic rather than linear.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSA.REI.B.4bReasoning with Equations and InequalitiesSolve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.
  • CCSS.MATH.CONTENT.HSA.REI.B.4aReasoning with Equations and InequalitiesUse the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)² = q that has the same solutions. Derive the quadratic formula from this form.
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