If you have ever seen the quadratic formula, you may think it has nothing to do with you or your life, but you would be very wrong. Many real world scenarios use quadratic equations, though we may not think of them when we solve the problem. For those that are completely new to algebra, Udemy offers a great beginner’s course in Algebra that can acquaint yourself with this higher level of mathematics.

A quadratic equation looks like this in standard form: *x ^{2} – 4x – 7 = 0*. Quad means squared, so we know that the quadratic equation must have a squared term, or it isn’t quadratic. The other numbers and terms are needed as well, but if you didn’t have a squared term, it would be a simple, but different, algebra problem.

**When You May Use Quadratic Equations**

If you shoot an arrow, fire a missile or throw stones or balls, you may want to know where it will go and how long it will take to get there. A quadratic equation can provide you with these answers. You can also use a quadratic equation to tell you how much to price a product and how many of those products to make, to tell how fast a vehicle is going, to study curved mirrors and lenses, and much more.

There are many reasons you might need a quadratic equation. Many people do these equations all the time and don’t realize they are doing them. This algebra is generally taught in middle or high school, though it is often also taught when one take college or university level Algebra and mathematics. Whether or not you will learn more about quadratic equations in college will be based on your college major; however, most colleges require basic algebra or a higher math, so unless you can test out of these courses, you will probably see quadratic equations in your college math course.

**Ways to Solve Quadratic Equations**

The most popular way to solve quadratic equations is to use a quadratic formula. This formula is: *-b ±√b ^{2} – 4ac/2a*

To solve, you will need to find the values of a, b, and c using the equation you are provided.

Another way to solve quadratic equations is to use the factoring method. To do this, set your equation equal to zero, so your equation looks like this: *ax ^{2} + bx + c = 0*. The first step would be to create a factor chart for the value of c. Once you have all of the factor pairs, you will need to do trial and error until you find a pair that adds up to the value of b. If this does not exist, you cannot use the factoring method. If you do find a pair that matches “b”, create two small equations and solve for “x”. Factoring is a very easy way to find your answer, though it doesn’t work for all problems. If it doesn’t, you will need to either complete the square or use the quadratic formula above.

If you need to complete the square in order to solve a quadratic equation, you must square a binomial and produce a perfect square trinomial. For example, if we have

(x + 3)^{2}, the perfect square trinomial would be: *x ^{2} + 6x + 9*. In essence, you will complete the square to find the last number so that you can work the quadratic equation using the quadratic formula.

**Hidden Quadratic Equations**

In some cases, your quadratic equation will not be written in the normal manner, and they will usually never be written in standard form. In this case you will need to format your equation into the normal method and then put it in standard form.

**Examples of Hidden Quadratic Equations**

There are many different ways of using quadratic equations. You will likely need a lot of practice, so consider this practical Algebra course on Udemy to help get you started. For now, let’s have a look at some more examples of what you are working with.

x^{2} = 4x – 3

2(w^{2} – 2w) = 7

z(z + 1) = 4

All of the above are quadratic equations, though they aren’t in standard form. Putting them in standard form and understanding which values are a, b, and c will help you solve the problem.

Hidden Quadratic | Standard Form | Values of A, B, C |

x^{2} = 4x – 3 | x^{2} – 4x + 3 = 0 | A = 1, B = -4, C = 3 |

2(w^{2} – 2w) = 7 | 2w^{2} – 4w – 7 = 0 | A = 2, B = -4, C = -7 |

z(z + 1) = 4 | z^{2} – z – 4 = 0 | A = 1, B = -1, C = -4 |

The first example is pretty straightforward; it is already in the normal method, so you just need to put it in standard form. However, the second example is not in the normal method. First, you must simplify by multiplying the “2” outside the bracket with both the squared w and the 2w. Then you would need to put the equation in standard quadratic form. Again, the third example is not in the normal method. You have to first simplify by multiplying the outside “z” with the inside z and make z squared, then multiply the outside z with 1 to get z. The last step is to put the equation in standard form.

**Examples of When You Will Need a Quadratic Equation and How to Solve Them**

There are many reasons you may need to use a quadratic equation; these three examples are just a very small sampling of when you may need them, how to understand why it will be a quadratic equation, and how to solve it.

**Factoring Example**

If you want to throw a ball in the air and want to know when it will hit the ground, you can use a quadratic equation. Let’s say that you throw a ball straight into the air from four meters above the ground and it has a velocity of 16 m/s. When will it hit the ground?

For this case, we must ignore air resistance, and then we need to find our equation. We know the height begins at four meters and that it travels up at 16 meters per second (19t). It is common knowledge (for most) that gravity will pull anything down, which will change the speed by approximately 5 meters per second per second (5t^{2}).

Therefore, our equation would look like this: *h = 4 + 19t = 5t ^{2}*. In this instance “t” is time and because both of our times are in m/s, “t” can be used for both measures of time.

When the height is zero, we know that our ball has finally reached the ground, which would make the equation: *4 + 19t – 5t ^{2} = 0*. Putting it into standard form, we will get the equation:

*-5t*and it will be even easier if we multiply everything by -1, so that our squared term is positive, giving us:

^{2}+ 19t + 4 = 0,*5t*.

^{2}– 19t – 4 = 0We must factor it using the method: **a x c, then add to give b**.

a x c = -20 and b = -19. The positive factors of -20 are 1, 2, 4, 5, 10, 20 and one factor must be negative.

We can easily find that -20 and 1 work (-20×1 = -20, and -20+1 = -19)

Rewrite the equation to be 5t^{2} – 20t + t – 4 = 0 and then factor the first and last two to make it: *5t(t – 4) + 1(t – 4) = 0*. Because there is a common factor (t – 4), we can write the equation as: *(5t + 1)(t – 4) = 0*.

The two solutions would be *5t + 1 = 0 or t – 4 = 0*. Solve for “t” in both equations:

5t + 1 = 0 t – 4 = 0

5t – 1 = 0 – 1 t + 4 = 0 + 4

5t = -1 t = 4

T = -1/5 or -0.2

Because time cannot be negative, we can eliminate the first solution, causing our ball to touch the ground after four seconds.

**Completing the Square Example**

If you wanted to find the dimensions of the base of the Temple Olympeion, you could find the answer by using complete the square processes. First, you need to know that the base of the temple is rectangular, and has a perimeter of 300 meters. You also need to know that the area of the base is 4400 m^{2}.

With this information, we can write a quadratic equation. First, we must know that the perimeter is equal to 2(length + width). Put in our perimeter of 300 meters and solve:

300 meters = 2 (L + W). Divide each side by 2 to get: 150 = L + W. If length is “x” and we are solving for width, we would have: width = (150 – x) meters.

We also know that area equals length multiplied by width.

Knowing that our length is “x” and our width is (150 – x), and the total area, we can write a quadratic equation:

Area equals length x width: Area = x (150 – x)

4400 = 150x – x^{2}. If we rewrite this in standard quadratic formula, it will look like this:

-x^{2} + 150x – 4400 = 0.

Next, we need to make the squared term positive, so multiply everything by -1:

X^{2} – 150x + 4400 = 0.

Move the constant to the right side:

X^{2} – 150x = -4400

Complete the square using (b/2)^{2}, where “b” is the x term and add it to both sides:

(b/2)^{2} = (-150/2)^{2} = (-75)^{2} = 5,625

X^{2} – 150x + 5,625 = -4400 + 5,625. Then simplify:

x^{2} – 150x + 5,625 = 1,225, to finish with this:

(x – 75)^{2} = 1,225. Take the square root of both sides:

x – 75 = ±√1,225, which equals: x – 75 = ±35. Solve for both negative and positive 35:

x – 75 = 35 x – 75 = -35

x = 35 + 75 x = -35 + 75

x = 110 x = 40

Because width equals (150 – x), we need to decide which of our “x” terms is width and which is length. Because 150 – 110 gives us 40, we know that width is 40 and length is 110 meters.

**Quadratic Formula Example**

Most people have a tendency to use the quadratic formula when solving quadratic equations, and the above two examples could be done using the quadratic formula, as well. However, sometimes it can be easier or more justifiable to use a different method.

Let’s say that you want to go on a three-hour river cruise that goes upstream and then back downstream. You want to know how fast the boat goes and how long the upstream journey will take, so that you know if your family will need extra-thick hats for warmth and so you know how many pictures to take going upstream versus back downstream.

First, you should know how far upstream the cruise takes you, which in this case will be 15 km. Next, you should know that the river you will be riding on has a current of 2 km/hr.

To begin, there are two actual speeds you need to consider. These include the speed relative to land, and the speed the boat makes moving in the water. For this scenario, let “x” be the boat’s speed in water and “v” be the boat’s speed relative to land.

The river flows downstream at 2 km/hr, so you will need to figure both upstream flow and downstream. When you go upstream, your speed will be reduced by 2 km/hr, as shown by v = x – 2; when you go downstream, your speed will be increased by 2 km/hr, as shown by v = x + 2.

These speeds can be turned into time by using the formula: time equals distance divided by speed. We also know that the total time of the cruise is three hours. This can be written as a formula as well: total time = time upstream plus time downstream, which equals three hours.

Put this all together to give yourself an equation:

Total Time = 15 / (x – 2) + 15 / (x + 2) = 3 hours. Solve for x (or in this case, simplify as much as you can:

Because you have fractions (or division), you will want to multiply them on both sides to remove the fraction, giving you: 15(x – 2) + 15(x + 2) = 3(x – 2) (x + 2). Simplify this to:

15x – 30 + 15x + 30 = 3 (x^{2} – 4). Switch everything around so that the squared term is first, and simplify:

3x^{2} – 12 = 30x; put the equation into the standard quadratic equation:

3x^{2 }– 30x – 12 = 0.

Next, we will use the quadratic formula, which is x = {-b ± √b^{2} – 4ac} / 2a. It is important to realize that a, b, and c come from the standard form of the quadratic equation. Plug the numbers into the quadratic formula and solve:

X = {-(-30) ±√((-30)^{2}-4x3x(-12))}/(2×3)

X = {30 ±√(900 + 144)} / 6

X = {30 ±√1044)} / 6

X = {30 ± 32.31} / 6

X = 30 + 32.31 / 6 and x = 30 – 32.31 /6

X = 10.39 x = -0.39

Because speed cannot be a negative number, we know that x = 10.39, meaning the speed of the boat is 10.39 km/hr. Therefore, you can determine that the boat will be going quite slowly, and you and your family probably won’t need very warm hats.

You have also determined the upstream and downstream journey times: the upstream journey is 15 / (10.39 – 2), which equals 1.79 hours or one hour and 47 minutes. The downstream journey is 15 / (10.39 + 2), which equals 1.21 hours, or one hour and 13 minutes. Based on this, you will have more time to take pictures while going upstream than on the return trip.

**Careers That Use Quadratic Equations**** **

The examples above should help you understand that you can use the quadratic equation in your own life. Yes, it can seem extremely difficult to do these equations, but once you see the pattern, memorize the formulas, and practice, it is quite easy to work out these problems with a calculator.

The above examples are meant to help you realize that you could actually use the quadratic formula and quadratic equations, if you so choose. However, the majority of people would not feel the need to find out how fast a boat is going or want to know the time it will take on each part of the journey. In some cases, your career could require the use of quadratic equations. These include military, law enforcement, engineering, science, management, and agricultural jobs.

The military generally uses quadratic equations to help them target objects that are flying through the air. It is important to pinpoint where their artillery will fall or hit their target, so they use quadratic equations. This can be used for airplanes, bullets, missiles, and tanks.

Police will use quadratic equations to figure out the trajectory of bullets and the speeds of cars. These are two very important things: if someone has been hurt by a bullet, the police need to know where the bullet was fired from and how fast it was going. This will help them figure out what type of gun was used and where the suspect was when they fired the gun. It is also important to be able to tell how fast a car was going, especially if there was an accident.

Engineers also use quadratic equations, along with many other types of advanced mathematics. If they are designing an object with a curve, for example, a quadratic equation may be used to ensure that the piece is made properly. Almost every type of engineer uses quadratic equations, including automotive engineers, electrical, chemical, audio, and computer engineers.

Many of the sciences also use quadratic equations, including astronomers, chemists and physicists. Additionally, agriculturists use these equations to produce bigger fields with the materials they are given, as well.

Several people don’t think of managers and clerical staff requiring quadratic equations, but in some cases, they must. For example, production and engineering managers must know how to do these equations, because they have to check the work of the engineer or production employee that did the actual equation. If the work is incorrect and is not noticed by the manager, the product will be made incorrectly.

Quadratic equations can also be used to determine how long products such as household appliances will last. These appliances become less safe over time, and it is important for manufacturers to have an idea of how long these appliances will work properly, so that they may include this information in user manuals.

The quadratic equation can also help developers of products. If you created a new product, you would want to know the type of profit you can expect. By using a quadratic equation, you can ultimately figure out your profit based on the amount of money used to make and advertise the product, along with how many units you make.

The same concept is true if you want to, for example, create a product that is twenty percent larger than it was previously. You have probably in the past seen a manufacturer claim that a product now offers 20 percent more, or is 15 percent larger. Most people don’t consider how the manufacturer determines the new size of the box or container – quadratic equations can play a big part.

Property surveyors must also use the quadratic equation, if they need to know the area of a property. If every piece of land was a perfect square, there wouldn’t be any trouble, but property isn’t divided into perfect squares.

Quadratic equations are needed in the lives of everyone; your life has been touched by the quadratic equation, even if you don’t realize it. Whether you prefer to use the quadratic formula or prefer the factoring method, you should learn to do quadratic equations; they may not be fun, but it can be satisfying to work the problem correctly and get the right answer.

There are many other types of mathematics courses available on Udemy if you would like to supplement your learning of mathematics. For example, Udemy offers this great Calculus I basics course that will help you learn the fundamentals of this higher math. And if you would like to further your understanding of Algebra beyond the quadratic equation, Udemy’s Advanced Algebra course can teach you everything you need to know or success.