# x2-11x+28=0: How To Solve Quadratic Equation [Easy Method]

Do quadratic equations complexity fascinate you, but with multiple solutions for the equation, it feels complex to figure out the easiest method?

If yes, you have covered all with the easiest way to solve x2-11x+28=0!

Factoring the equation, using the quadratic formula, and completing the square are some approaches to solving the problem.

So, without further ado, let’s dig into the easiest way to solve the equation.

## Which Are The Methods To Solve x^2-11x+28=0?

The methods to solve the quadratic equation x²-11x+28=0 include factorizing expression, completing the square, or using graphical methods.

This is because of the number of assumptions involved.

Therefore, it’s better to use a full-proof quadratic equation method instead of a hit-and-trial, time-consuming way to rule out the solution.

## How To Solve x2-11x+28=0 Easily?

**To solve x²-11x+28=0, use the easiest method by substituting the values of a=1, b=-11, and c=28 in the quadratic formula x=-b±√(b²-4ac)/2a, returning values of x=7 and 4. **

### Step 1: Check If It Is In The Form Of ax²+bx+c=0

To solve a quadratic equation, it must be ax²+bx+x=0; if it is not, then rearrange the equation. To do so, shift the terms in the LHS by keeping the like-minded terms together.

x²-11x+28=0 is already in the form of ax²+bx+c=0.

### Step 2: Find The Values of a, b and c

In the formula of ax²+bx+c, the values of a, b, and c according to the given equation are a=1, b=-11, and c=1.

### Step 3: Find The Value Of The Discriminant

The discriminant determines the number of solutions or, to be precise, the roots of the quadratic equation. It is also used to substitute in the quadratic formula.

To find the value of the discriminant, use the formula b²-4ac.

According to the equation, the value of the discriminant value will be:

- b²-4ac
- (-11)²-2(1)(28)

Solving the square;

- 121-2(1)(28)

Multiplying the brackets;

- 121-1129

Subtracting The terms;

- 9

9 is the value of the discriminant, using which we can evaluate the roots and find the solution of the quadratic equation.

### Step 4: Evaluate the Roots of the Equation

The roots of the quadratic equation find the type of roots and solutions available.

**If b²-4ac>0, the equation has two real and unique solutions.**

- If b²-4ac=0, the equation will have only one real solution.

- If b²-4ac<0, the equation will have imaginary roots.

Here, the value of b²-4ac=9, which is greater than 0; hence, the equation has two real and unique solutions.

### Step 5: Substitute The Value of Discriminant In The Quadratic Formula

Now, the next crucial step of implementation will be using the quadratic formula,

- x= -b±√(b²-4ac)/2a

Substituting the values of a, b, and discriminant from the above formula;

- x=-(-11)±√(9)/2(1)

The result of multiplying (-)(-) is always (+), hence solving the first term;

- x=11±√(9)/2(1)

Finding the square root of 9;

- x=11±3/2

The ± indicates that the operation can be both (+) and (-), and thus the equation will have two distinct values of x. Besides, the discriminant already states that it will have two real and unique solutions that comply with the formula.

### Step 6: Separate the Equations And Find The Values Of x

Since the operation performed can be positive and negative, and x will have two values, you have to implement both operations.

Let’s begin with using the (+)operation using the formula x=-b+√(b²-4ac)/2a;

Since the value of x as per the equation is 11+3/2, substituting it in the formula;

- x=11+3/2
- x=14/2
- x=7

Similarly, let’s use the (-) operation and use the formula x=-b-√(b²-4ac)/2a, whose value is 11-3/2

- x=11-3/2
- x=8/2
- x=4

Therefore, the values of x are now known;

x=7 and x=4.

## Frequently Asked Questions

**What is the quadratic formula?**

The Quadratic formula is for the equations, which are in the form of ax²+bx+c=0.

The formula is used like x=-b±√(b²-4ac)/2a where b²-4ac is the value of the discriminant

**What are the solutions to x2-11x+28=0?**

The solutions to the equation x²-11x+28 are x=7 and x=4.

**How can I evaluate the solutions of a quadratic equation? **

To evaluate the solutions of a quadratic equation, substitute the value of x you found out and check if it equals the RHS.

**What are the ways to solve quadratic equations?**

To solve the quadratic equations, you can use methods like factoring the equation, completing the square, using the quadratic formula, or following the graphical way.

## Conclusion

The easiest method to solve x2-11x+28=0 is by using the quadratic formula. You must determine the values of a, b, and c, which are 1, -11, and 28, to use the quadratic equation. Further, check the roots of the quadratic equation by finding the value of discriminant b²-4ac. The value is 3, which is greater than 0; hence, the equation has two real and unique solutions.

Next, substitute it in the formula x=-b±√(b²-4ac)/2a and separate the equations as x=-b+√(b²-4ac)/2a and x=-b+√(b²-4ac)/2a. Upon substitution, the values come out to be x=7 and x=4.