# The Quadratic Equation Formula: 3 Methods for Solving The quadratic equation formula is used in algebra and can seem a little daunting at first because the formula itself is fairly complex compared with others one might have seen. However, it is quite easy to use the formula once you understand it. There are three different ways to solve quadratic equations. The first includes factoring, though not every math problem can be factored. The second is to complete the square, and the third way to solve these problems is to use the quadratic formula. There are many courses from Udemy that can help you learn in-depth algebra easily, but here is a quick overview of quadratic equations and the quadratic equation formula.

The quadratic equation looks like this: x2 – 3x + 2 = 0. The x is the unknown number and makes an equation that can be solved easily using one of the three ways—factoring, quadratic formula and squares.

## Factoring

The first step in factoring is to combine all of the like terms and keep them all to one side in the equation. You must keep the x2 term positive, as well.

Consider 2x2 – 8x – 4 = 3x – x2

In the above equation, you must solve for x, but first you must move all the like terms together and keep the squared term positive. To do this, place all the squared terms together, then group the x terms together, and then include any other numbers and make it equal to zero. Remember to add or subtract correctly when moving from one side of the equal sign to another.

Therefore, your equation would look like this: 2x2 + x2– 8x – 3x – 4 = 0. Next, you would group these terms together more, if possible. For example, the squared terms can be added together and the x terms can also be added, giving you this equation:

3x2 – 11x – 4 = 0.

Next, you must factor the expression. To do this, use factors of the x2 term, which is three, and the constant term, which is negative four, then multiply them and add to the middle term.

Because 3x2 has only one set of factors (3x and x), write them in parentheses, which would look like this: (3x + 1) (x – 4) = 0.

Next, using the process of elimination, plug in the factors of four until you find a combination that makes -11x when multiplied. The factors one and four will work to create -11x, so there is no other work that you must do. When you multiply (3x + 1) (x – 4), you get -3x2 – 12x + x – 4. When you simplify further, you get a middle number of -11x, which is what you needed.

Next, you would set each of the parentheses to zero and solve. This would mean that

3x + 1 = 0 and x – 4 = 0. Solving each independently would give you:

3x + 1 = 0                               x – 4 = 0

– 1 = 0 – 1                             – 4 = 0 – 4

3x = -1                                    x = -4

3x/3 = -1/3

X = -1/3

This means that x equals -1/3 and -4.

4x2 – 5x – 13 = x2 – 5, you would move all the like terms to one side of the equation and set the other side to zero, which would make the equation look like this:

4x2 – x2 – 5x – 13 + 5 = 0, then you would simplify: 3x2 –5x – 8 = 0.

Write the problem using the quadratic formula, which is x = {-b +/-√(b2 – 4ac)}/2a. Before you can do this, you must decide the values of a, b, and c. In every case, the x2 is a, the x term is b and the constant term is c.

With the equation 3x2 –5x – 8 = 0, a = 3, b = 5, and c = 8.

Substitute the values of a, b, and c into the quadratic equation:

{-b +/-√(b2 – 4ac)}/2a would turn into:

{-(-5) +/-√((-5)2 – 4(3)(-8))}/2(3). Simplify this as much as possible to finish with:

{-(-5) +/-√((25) – (-96))}/6; simplify further to update positive and negatives and multiply all remaining terms: {5 +/-√(121)}/6

Next, you must simplify the square root. When you receive a whole number, there is nothing else to do, but if the square root is not perfect, then you will simplify to the simplest radical version. However, the square root of 121 is 11, so it is perfect.

{5 +/-√(121)}/6 will get simplified to x = (5 +/- 11)/6. Next, you will solve for the positive and negative, meaning that:

X = (5 + 11)/6 and X = (5 – 11)/6

Solve and simplify if necessary:

X = 16/6 or 8/3 and X = -6/6 or -1, so that x = (-1, 8/3)

## Completing the Square

The last way to solve a quadratic equation is to complete the square. Again, you will start by moving all the terms to one side and equating it to zero.

In the problem 2x2 – 9 = 12x, you would write it like this: 2x2 – 12x -9 =0.

Make a note that the “a” term is 2, the b term is -12, and the c term is -9.

Next, move the c term to the other side, giving you 2x2 – 12x = 9.

Divide both sides by the “a” term, to give you x2/2 – 12x/2 = 9/2 and simplify to make it:

x2 – 6x = 9/2.

Your next step is to divide the “b” term by two, square it, and add it to both sides. For this problem, the “b” term is -6. Therefore:

-6/2 = -3; (-3)2 = 9. This means the equation will now look like this:

x2 – 6x + 9 = 9/2 +9.

To simplify, you must factor the constant terms on the left, which gives you (x – 3) (x – 3). Add the terms on the right of the equal sign as well, which leaves you with:

(x – 3)2 = 9/2 + 18/2. Once you simplify, you are left with (x – 3)2 = 27/2.

Find the square root of both sides. Since the left side is squared, you can simplify remove the square, and that is the square root, and write the square root of 27/2 as ±√27/2. Therefore, you have an equation that now looks like this:

X – 3 = ±√27/2.

Simplify the radical, and then solve for x.

To simplify ±√27/2, find the perfect square for the numbers 27 and 2 or their factors.

9 x 3 = 27 and 9 is a perfect square. Therefore, take 9 out of the radical sign; to do this, remove the number 9 from the radical and write in the number 3, its square root, outside the radical sign.

You will also want to add three to both sides of the equation to get “x” by itself:

X = 3 ± 3√6/2. Simplify further to get:

X = 3 + (√6)/2 and      x = 3 – (√6)/2, giving you:

X = (4.22, 1.78)

Knowing how to use the quadratic equation formula, as you can learn from various Algebra courses offered by Udemy, can help to make math much easier to understand. Take a look at all the math courses from Udemy, such as this great beginner Algebra course that will give you a strong foundation in this complex type of math. The skills that you learn will be invaluable to you in many ways.