Solve Quadratic equations x2+x-756=0 Tiger Algebra Solver (2025)

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1Factoring x2+x-756

The first term is, x2 its coefficient is 1.
The middle term is, +x its coefficient is 1.
The last term, "the constant", is -756

Step-1 : Multiply the coefficient of the first term by the constant 1-756=-756

Step-2 : Find two factors of -756 whose sum equals the coefficient of the middle term, which is 1.

-756+1=-755
-378+2=-376
-252+3=-249
-189+4=-185
-126+6=-120
-108+7=-101
-84+9=-75
-63+12=-51
-54+14=-40
-42+18=-24
-36+21=-15
-28+27=-1
-27+28=1That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -27 and 28
x2 - 27x+28x - 756

Step-4 : Add up the first 2 terms, pulling out like factors:
x•(x-27)
Add up the last 2 terms, pulling out common factors:
28•(x-27)
Step-5:Add up the four terms of step4:
(x+28)•(x-27)
Which is the desired factorization

Equation at the end of step 1 :

 (x + 28) • (x - 27) = 0 

Step 2 :

Theory - Roots of a product :

2.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

2.2Solve:x+28 = 0Subtract 28 from both sides of the equation:
x = -28

Solving a Single Variable Equation:

2.3Solve:x-27 = 0Add 27 to both sides of the equation:
x = 27

Supplement : Solving Quadratic Equation Directly

Solving x2+x-756 = 0 directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex:

3.1Find the Vertex ofy = x2+x-756Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,1, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the x-coordinate of the vertex is given by -B/(2A). In our case the x coordinate is -0.5000Plugging into the parabola formula -0.5000 for x we can calculate the y-coordinate:
y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 - 756.0
or y = -756.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2+x-756
Axis of Symmetry (dashed) {x}={-0.50}
Vertex at {x,y} = {-0.50,-756.25}
x-Intercepts (Roots) :
Root 1 at {x,y} = {-28.00, 0.00}
Root 2 at {x,y} = {27.00, 0.00}

Solve Quadratic Equation by Completing The Square

3.2Solvingx2+x-756 = 0 by Completing The Square.Add 756 to both side of the equation :
x2+x = 756

Now the clever bit: Take the coefficient of x, which is 1, divide by two, giving 1/2, and finally square it giving 1/4

Add 1/4 to both sides of the equation :
On the right hand side we have:
756+1/4or, (756/1)+(1/4)
The common denominator of the two fractions is 4Adding (3024/4)+(1/4) gives 3025/4
So adding to both sides we finally get:
x2+x+(1/4) = 3025/4

Adding 1/4 has completed the left hand side into a perfect square :
x2+x+(1/4)=
(x+(1/2))(x+(1/2))=
(x+(1/2))2
Things which are equal to the same thing are also equal to one another. Since
x2+x+(1/4) = 3025/4 and
x2+x+(1/4) = (x+(1/2))2
then, according to the law of transitivity,
(x+(1/2))2 = 3025/4

We'll refer to this Equation as Eq. #3.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x+(1/2))2 is
(x+(1/2))2/2=
(x+(1/2))1=
x+(1/2)

Now, applying the Square Root Principle to Eq.#3.2.1 we get:
x+(1/2)= 3025/4

Subtract 1/2 from both sides to obtain:
x = -1/2 + √ 3025/4

Since a square root has two values, one positive and the other negative
x2 + x - 756 = 0
has two solutions:
x = -1/2 + √ 3025/4
or
x = -1/2 - √ 3025/4

Note that 3025/4 can be written as
3025 / √4which is 55 / 2

Solve Quadratic Equation using the Quadratic Formula

3.3Solvingx2+x-756 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B and C are numbers, often called coefficients, is given by :

-B± √B2-4AC
x = ————————
2A
In our case,A= 1
B= 1
C=-756
Accordingly,B2-4AC=
1 - (-3024) =
3025
Applying the quadratic formula :

-1 ± √ 3025
x=——————
2
Can 3025 be simplified ?

Yes!The prime factorization of 3025is
5•5•11•11
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

3025 =√5•5•11•11 =5•11•√ 1 =
±55 •√ 1 =
±55

So now we are looking at:
x=(-1±55)/2

Two real solutions:

x =(-1+√3025)/2=(-1+55)/2= 27.000

or:

x =(-1-√3025)/2=(-1-55)/2= -28.000

Two solutions were found :

  1. x = 27
  2. x = -28
Solve Quadratic equations x2+x-756=0 Tiger Algebra Solver (2025)

FAQs

Which are the solutions of the quadratic equation x2 7x 4 7 07 0? ›

x2 = 7x + 4. -7, 0, 7, 0. Therefore, the solution is x = [7 ± √65]/2.

How is the quadratic formula derived? ›

The quadratic formula, x=- b±sqrt(b^2-4ac)/2a, can be derived by completing the square on the general standard form a quadratic equation. Recall that completing the square is a method for solving quadratic equations.

What are the 4 methods in finding the solution of the quadratic equation? ›

Answer: There are various methods by which you can solve a quadratic equation such as: factorization, completing the square, quadratic formula, and graphing. These are the four general methods by which we can solve a quadratic equation.

What is the solution set for the equation x2 7x 12 0? ›

x = 3, x = 4.

What are the 4 steps to solve a quadratic equation? ›

Solving Quadratic Equations
  • Put all terms on one side of the equal sign, leaving zero on the other side.
  • Factor.
  • Set each factor equal to zero.
  • Solve each of these equations.
  • Check by inserting your answer in the original equation.

How to do a quadratic formula step by step? ›

Applying the Quadratic Formula

Step 1: Identify a, b, and c in the quadratic equation a x 2 + b x + c = 0 . Step 2: Substitute the values from step 1 into the quadratic formula x = − b ± b 2 − 4 a c 2 a . Step 3: Simplify, making sure to follow the order of operations.

How to solve a quadratic equation without a formula? ›

Set the equation equal to zero. If the quadratic side is factorable, factor, then set each factor equal to zero. If the quadratic equation involves a SQUARE and a CONSTANT (no first degree term), position the square on one side and the constant on the other side. Then take the square root of both sides.

How to solve a quadratic equation by factoring? ›

To solve an quadratic equation using factoring :
  1. Transform the equation using standard form in which one side is zero.
  2. Factor the non-zero side.
  3. Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero).
  4. Solve each resulting equation.

Does the quadratic formula always work? ›

Finally, the quadratic formula will work on any quadratic equation. However, if using the formula results in awkwardly large numbers under the radical sign, another method of solving may be a better choice.

How do you find roots in a quadratic equation? ›

Important Formulas for Quadratic Equation Roots include:

ax² + bx + c = 0 is a quadratic equation. Use the formula x = (-b ± √ (b² – 4ac) )/2a. to calculate the roots. D = b² – 4ac is the discriminant.

Which are the solutions of the quadratic equation? ›

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. The roots of any polynomial are the solutions for the given equation.

How many real solutions are there to the equation x2 − 7x 12 0? ›

👉Therefore, the solutions to the quadratic equation X² - 7x + 12 = 0 are x = 3 and x = 4.

How do you find out how many solutions a quadratic has? ›

If b2 - 4ac is positive (>0) then we have 2 solutions. If b2 - 4ac is 0 then we have only one solution as the formula is reduced to x = [-b ± 0]/2a. So x = -b/2a, giving only one solution. Lastly, if b2 - 4ac is less than 0 we have no solutions.

Which are the solutions of x2 13x 4 0 13 0 13? ›

Answer. The solutions of the equation are -12.68 and -0.31.

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