The beauty of the quadratic formula is that it **always** works for solving quadratic equations.

Just like we know the sun will rise in the morning tomorrow, we know the quadratic formula will provide the solutions (real and imaginary) to our quadratic equations. If you are looking for a little more beauty (and a lot less erasing) when solving quadratics, keep reading to review the quadratic formula, understand the derivation of the quadratic formula, and see examples of how to use the quadratic formula!

**The quadratic formula**: *as reliable as the sunrise*!

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What We Review

## What is the Quadratic Formula?

Let’s start with looking at the full quadratic formula below:

The letters a, b, and c come from the standard form of a quadratic equation:

The coefficient in front of x^2 is a, the coefficient in front of x is b, and the coefficient without a variable is c.

You can learn more about standard form and other forms of quadratic equations in our review article about the forms of quadratics.

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**History of the Quadratic Formula**

### Early History

Solving quadratic equations is no modern accomplishment. In fact, the ancient Babylonians were completing the square to solve quadratic equations long before the word “Algebra” even existed, according to Hackworth and Howland in the text Introductory College Mathematics: History of Real Numbers.

Imagine solving quadratic equations with an abacus instead of pulling out your calculator. According to Mathnasium, not only the Babylonians but also the Chinese were solving quadratic equations by completing the square using these tools.

In the year 700 AD, Brahmagupta, a mathematician from India, developed a general solution for the quadratic equation, but it was not until the year 1100 AD that the solution we know today was developed by another mathematician from India named Bhaskara, as stated by Mathnasium.

Hackworth and Howland let us know that until Hindu mathematics, mathematics developed in India, existed, numbers did not appear the way we know them, with a base 10 system. In other words, as these developments were happening to solve quadratic equations, many other developments occurred in mathematics. Famed ancient mathematicians, such as Pythagoras, did not understand the existence of numbers such as \sqrt{2} or \pi. In fact, as Mathnasium let’s us know, Bhaskara’s recognition of two solutions to the square root of a number helped him to solve quadratic equations. Britannica informs us that Bhaskara also was able to approximate the value of \pi as 3.141666.

### More Recent History

During the time period between Brahmagupta and Bhaskara, an Arab mathematician, Al-Khwarizmi, also solved quadratic equations, according to Mathnasium. He wrote a text titled Al-jabr wa’l muqabala from which we get the word algebra, according to Hackworth and Howland. Hackworth and Howland also let us know that in his text, Al-Khwarizmi explains with specific steps how to solve equations, including quadratic equations.During the Renaissance, around 1545 AD, Girolamo Cardano built upon the works of Al-Khwarizmi, as stated by Mathnasium, including imaginary solutions. While the works of Cardano include imaginary numbers, Al-Khwarizmi did not even include negative solutions, according to Hackwork and Howland.

Finally, in 1637, Rene Descartes included the quadratic formula as we know it today in his work La Geometrie, as told by Mathnasium.

## Deriving the Quadratic Formula

We can derive the quadratic formula by completing the square of the Standard Form of a quadratic equation. We will begin by setting the value of y equal to 0. Remember, when solving a quadratic equation, we are finding the zeros or x-intercepts:

y=ax^2+bx+c

0=ax^2+bx+c

ax^2+bx+c=0

Then, we will divide all values by a, ensuring there is no coefficient in front of x^2. To isolate the terms with the variable, x, we will move the constant, \dfrac{c}{a}, to the other side of the equation by subtracting \dfrac{c}{a} from both sides of the equation:

ax^2+bx+c=0

x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0

x^2+\dfrac{b}{a}x=-\dfrac{c}{a}

Now, to complete the square, we must determine half of the coefficient in front of x and square that value. This will be the value to add to both sides of our equation. The coefficient in front of x is \dfrac{b}{a}:

\dfrac{b}{a} \times \dfrac{1}{2} = \dfrac{b}{2a}

{(\dfrac{b}{2a})}^2=\dfrac{b^2}{4a^2}

Now, we will add \dfrac{b^2}{4a^2} to both sides of the equation:

x^2+\dfrac{b}{a}x=-\dfrac{c}{a}

x^2+\dfrac{b}{a}x +\dfrac{b^2}{4a^2}=-\dfrac{c}{a}+ \dfrac{b^2}{4a^2}

We will factor the left side because it is a perfect square. We will also simplify the right side of the equation. To solve for x we will need to take the square root of both sides and isolate x by subtracting \dfrac{b}{2a} from both sides. Remember that when we take the square root, we must include the positive and the negative solution:

x^2+\dfrac{b}{a}x +\dfrac{b^2}{4a^2}=-\dfrac{c}{a}+ \dfrac{b^2}{4a^2}

(x+\dfrac{b}{2a})^2=\dfrac{b^2-4ac}{4a^2}

x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}

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

From here, all we need to do is simplify the expression. When we evaluate the square root of the denominator, both denominators are the same, allowing us to combine the fractions:

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

**x= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} **

It took a bit of work, but we have now *proven* the quadratic formula will work in all cases!

If you’d like to see this process in visual form, below is a brief video from YouTuber patrickJMT showing how to derive the quadratic formula:

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**When to Use the Quadratic Formula**

Sometimes, when trying to solve a quadratic equation by factoring, we hit a block in the road. Not every quadratic equation is factorable.

The great news about the quadratic formula is that you may always use it! There are no quadratic equations where the quadratic formula will fail to provide a solution. Even in cases where there are no real solutions, the quadratic formula will still provide solutions! Check out our section below about the discriminant to learn more about how the quadratic formula always provides appropriate solutions.

However, when a quadratic equation is factorable, factoring is often a more efficient method to solve the equations. You can actually determine if a quadratic equation is factorable using something called the discriminant (for more info on this, jump to our section below on discriminants and factoring).

Quadratic equations can also be solved using completing the square, the method we used above to derive the quadratic formula.

Conclusively, you will never go wrong with the quadratic formula, but there are other ways to solve quadratic equations, too!

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## Quadratic Formula Examples with Answers (Step by Step)

### Real Solutions

Let us try for ourselves! We will solve the quadratic equation:

**y=2x^2+12x-1**

When we solve the quadratic equation, we are determining the zeros, or x-intercepts, so we make the value of y equal to 0.

First, we must identify the variables a, b, and c. Remember, the standard form of a quadratic equation is y=ax^2+bx+c.

We will compare our equation to the standard form of a quadratic equation to determine the values of a, b, and c.

\color{red}{2x^2} \color{blue}{+12x}\color{green}{-1}=0

\color{red}{ax^2} \color{blue}{+bx}\color{green}+c=0

We can see that \color{red}{a=2}, \color{blue}{b=12}, and \color{green}{c=-1}.

Now, we can substitute these values into the quadratic formula:

\dfrac{\color{blue}{-b} \pm \sqrt{\color{blue}{b^2} - 4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}

\dfrac{\color{blue}{(-12)} \pm \sqrt{\color{blue}{(12)^2} - 4\color{red}{(2)}\color{green}{(-1)}}}{2\color{red}{(2)}}

Now, we must simplify the expression.

\dfrac{\color{blue}{(-12)} \pm \sqrt{\color{blue}{(12)^2} - 4\color{red}{(2)}\color{green}{(-1)}}}{2\color{red}{(2)}}

\dfrac{(-12) \pm \sqrt{144 - (-8)}}{4}

\dfrac{(-12) \pm \sqrt{152}}{4}

\dfrac{(-12) \pm 2\sqrt{38}}{4} = \dfrac{(-6) \pm \sqrt{38}}{2}

Therefore, the two real solutions to the equation:

y=2x^2+12x-1

…are \dfrac{(-6) + \sqrt{38}}{2} and \dfrac{(-6) - \sqrt{38}}{2} .

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### Imaginary Solutions

Now, let us solve a quadratic equation with **imaginary** solutions. We will solve the quadratic equation:

y=7x^2-x+9

Just like before, when we solve the quadratic equation, we are determining the zeros, or x-intercepts, so we make the value of y equal to 0.

We will compare our equation to standard form, set equal to 0, to determine the values of a, b, and c.

\color{red}{7x^2}\color{blue}{-x}\color{green}{+9}=0

\color{red}{ax^2}\color{blue}{+bx}\color{green}+c=0

We can see that \color{red}{a=7}, \color{blue}{b=-1}, and \color{green}{c=9}.

Now, we can substitute these values into the quadratic formula.

\dfrac{\color{blue}{-b} \pm \sqrt{\color{blue}{b^2} - 4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}

\dfrac{\color{blue}{-(-1)} \pm \sqrt{\color{blue}{(-1)^2} - 4\color{red}{(7)}\color{green}{(9)}}}{2\color{red}{(7)}}

Now, we must simplify the expression.

\dfrac{\color{blue}{-(-1)} \pm \sqrt{\color{blue}{(-1)^2} - 4\color{red}{(7)}\color{green}{(9)}}}{2\color{red}{(7)}}

\dfrac{1 \pm \sqrt{1-252}}{14}

\dfrac{1 \pm \sqrt{-251}}{14}

Because we have the square root of a negative number, we have two imaginary solutions. We must rewrite \sqrt{-251} using imaginary numbers. Then, we will write our solution in complex number form a+bi.

\dfrac{1 \pm \sqrt{-251}}{14}

\dfrac{1 \pm i\sqrt{251}}{14}

\dfrac{1}{14} \pm \dfrac{i \sqrt{251}}{14}

Therefore, the two imaginary solutions to the equation y=7x^2-x+9 are

\dfrac{1}{14} + \dfrac{i \sqrt{251}}{14} and \dfrac{1}{14} - \dfrac{i \sqrt{251}}{14}.

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## Free Quadratic Formula Worksheet

Math teachers: are you looking for an easy-to-use quadratic formula worksheet that includes brief notes and some basic practice problems?

*Preview of quadratic formula worksheet:*

## What is the Discriminant of a Quadratic Equation?

Now that we have solved equations using the quadratic formula, determining the discriminant will seem simple! The discriminant is a piece of the quadratic formula.

The discriminant can give us useful information without doing all of the work to solve for the zeros of a quadratic equation. There are three possible outcomes:

- The discriminant is
**positive**: b^2-4ac>0 - The discriminant is
**zero**: b^2-4ac=0 - The discriminant is
**negative**: b^2-4ac<0

In each case, we can determine something about what type of solution we will have. Remember, in the quadratic formula, the discriminant is the part **under the square root**.

- If the discriminant is
**positive**, this means we are taking the square root of a positive number. We will have a positive and negative real solution. This equation will have two real solutions, or x-intercepts.

- If the discriminant is
**zero**, we are taking the square root of zero, which is zero. Adding zero and subtracting zero produce the same value, so the equation will have only one x-intercept.

- Finally, if the discriminant is
**negative**, we are taking the square root of a negative number. This equation will have no real solutions, but two imaginary solutions because i=\sqrt{-1}.

This table summarizes how to use the discriminant to determine the number and type of solutions in a quadratic equation:

Discriminant | Number of Solutions to Equation |
---|---|

Positive | Two Real Solutions |

Negative | Two Imaginary Solutions |

Zero | One Real Solution |

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### QUICK TIP: Discriminants and Factoring

You can also use the discriminant to determine whether or not equations are factorable. If a, b, and c are integer values, when the discriminant is a perfect square, the quadratic equation is factorable.

In other words, when a, b, and c are integer values:

If b^2-4ac is a perfect square, ax^2+bx+c is factorable.

This is explained further in The University of Chicago School Mathematics Project Algebra textbook, Vol. 2 including great examples of factorable equations on this video.

**How to Find the Discriminant of a Quadratic Equation**

### Positive Discriminant

Let us determine the discriminant of the quadratic equation:

**y=8x^2-13x+1**

First, we must determine the values of a, b, and c. We will compare our equation to standard form, which is y=ax^2+bx+c.

y=\color{red}{8x^2}\color{blue}{-13x}\color{green}{+1}

y=\color{red}{x^2}\color{blue}{+bx}\color{green}+c

We can see that \color{red}{a=8}, \color{blue}{b=-13}, and \color{green}{c=1}.

Now, we can substitute these values into the discriminant and simplify.

\color{blue}{b^2}-4\color{red}{a}\color{green}{c}

\color{blue}{(-13)^2}-4\color{red}{(8)}\color{green}{(1)}

169-32

137

We know that 137>0. Since the discriminant is **positive**, we know the equation y=8x^2-13x+1 must have **two real solutions**. We can see that the quadratic equation y=8x^2-13x+1 crosses the x-axis twice by graphing the equation.

### Negative Discriminant

Now, let us determine the discriminant of a different quadratic equation:

**y=-x^2+5x-7**

We will first determine the values of a, b, and c.

We will compare our equation to standard form, which is y=ax^2+bx+c.

y=\color{red}{-x^2}\color{blue}{+5x}\color{green}{-7}

y=\color{red}{x^2}\color{blue}{+bx}\color{green}+c

We can see that \color{red}{a=-1}, \color{blue}{b=5}, and \color{green}{c=-7}.

Now, we can substitute these values into the discriminant and simplify.

\color{blue}{b^2}-4\color{red}{a}\color{green}{c}

\color{blue}{(5)^2}-4\color{red}{(-1)}\color{green}{(-7)}

25-28

-3

We know that -3<0. Because the discriminant is **negative**, we know the quadratic equation y=-x^2+5x-7 has **two imaginary solutions**. We can see on the graph that the function does not cross the x-axis.

### Discriminant of Zero

Lastly, let us determine the value of the discriminant for the quadratic equation:

**y=x^2-6x+9**

We begin by determining the values of a, b, and c.

We will compare our equation to standard form, which is y=ax^2+bx+c.

y=\color{red}{x^2}\color{blue}{-6x}\color{green}{+9}

y=\color{red}{x^2}\color{blue}{+bx}\color{green}+c

We can see that \color{red}{a=1}, \color{blue}{b=-6}, and \color{green}{c=9}.

Now, we can substitute these values into the discriminant and simplify.

\color{blue}{b^2}-4\color{red}{a}\color{green}{c}

\color{blue}{(-6)^2}-4\color{red}{(1)}\color{green}{(9)}

36-36

We know that 0=0. The discriminant with a **value of 0** tells us that the equation has only **one x-intercept**. We can see the function hit the x by graphing this quadratic equation.

*More of a visual learner?* Watch the great video explanation below on how to find the discriminant of a quadratic:

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## The 5 Best Quadratic Formula Songs

What better way to memorize a formula than to listen to an annoying song? Yes, there are some pleasant songs out there, but the song that will play in your head even when you wish it would stop is the song that will best help you to memorize the formula. Here are some great options for you!

- With over three million views, we had to start with the all-original quadratic formula song.
- Here’s another original version of a quadratic formula song — warning: this might get stuck in your head.
- Here is a great cover of a One Direction song using the quadratic formula.
- While this rap may not get the formula stuck in your head, it does review three methods for solving quadratic equations!
- To round out our list, one more cover. The chorus on this version is a truly enjoyable way to sing the quadratic formula (starts at 1:00).

## Practice with Quadratic Formula

For practice questions focused on the quadratic formula, explore Albert’s Algebra 1 practice course! All Albert questions include explanations of solutions and how to avoid common mistakes. Math teachers: read our comprehensive guide on how to teach quadratic equations including strategies, real-life examples, and prerequisite student skills.

Additionally, licensed Albert teachers can assign students this short Algebra 1 Topic Quiz that focuses on solving quadratic equations.

Finally, check out our other detailed Algebra 1 review guides to learn more about quadratics.