Solution - Quadratic equations

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "n2"   was replaced by   "n^2". 

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  n²+301n-4740 

The first term is,  n²  its coefficient is  1 .
The middle term is,  +301n  its coefficient is  301 .
The last term, "the constant", is  -4740 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -4740 = -4740 

Step-2 : Find two factors of  -4740  whose sum equals the coefficient of the middle term, which is   301 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -15  and  316 
                     n² - 15n + 316n - 4740

Step-4 : Add up the first 2 terms, pulling out like factors :
                    n • (n-15)
              Add up the last 2 terms, pulling out common factors :
                    316 • (n-15)
Step-5 : Add up the four terms of step 4 :
                    (n+316)  •  (n-15)
             Which is the desired factorization

Equation at the end of step  1  :

  (n + 316) • (n - 15)  = 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 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 2.2      Solve  :    n+316 = 0 

 Subtract  316  from both sides of the equation : 
                      n = -316

Solving a Single Variable Equation :

 2.3      Solve  :    n-15 = 0 

 Add  15  to both sides of the equation : 
                      n = 15

Supplement : Solving Quadratic Equation Directly

Solving    n2+301n-4740  = 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 :

 3.1      Find the Vertex of   y = n²+301n-4740

Parabolas 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,An²+Bn+C,the  n -coordinate of the vertex is given by  -B/(2A) . In our case the  n  coordinate is  -150.5000  

 Plugging into the parabola formula  -150.5000  for  n  we can calculate the  y -coordinate : 
  y = 1.0 * -150.50 * -150.50 + 301.0 * -150.50 - 4740.0
or   y = -27390.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = n²+301n-4740
Axis of Symmetry (dashed)  {n}={-150.50} 
Vertex at  {n,y} = {-150.50,-27390.25} 
 n -Intercepts (Roots) :
Root 1 at  {n,y} = {-316.00, 0.00} 
Root 2 at  {n,y} = {15.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 3.2     Solving   n²+301n-4740 = 0 by Completing The Square .

 Add  4740  to both side of the equation :
   n²+301n = 4740

Now the clever bit: Take the coefficient of  n , which is  301 , divide by two, giving  301/2 , and finally square it giving  90601/4 

Add  90601/4  to both sides of the equation :
  On the right hand side we have :
   4740  +  90601/4    or,  (4740/1)+(90601/4) 
  The common denominator of the two fractions is  4   Adding  (18960/4)+(90601/4)  gives  109561/4 
  So adding to both sides we finally get :
   n²+301n+(90601/4) = 109561/4

Adding  90601/4  has completed the left hand side into a perfect square :
   n²+301n+(90601/4)  =
   (n+(301/2)) • (n+(301/2))  =
  (n+(301/2))²
Things which are equal to the same thing are also equal to one another. Since
   n²+301n+(90601/4) = 109561/4 and
   n²+301n+(90601/4) = (n+(301/2))²
then, according to the law of transitivity,
   (n+(301/2))² = 109561/4

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
   (n+(301/2))²   is
   (n+(301/2))^2/2 =
  (n+(301/2))¹ =
   n+(301/2)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
   n+(301/2) = √ 109561/4

Subtract  301/2  from both sides to obtain:
   n = -301/2 + √ 109561/4

Since a square root has two values, one positive and the other negative
   n² + 301n - 4740 = 0
   has two solutions:
  n = -301/2 + √ 109561/4
   or
  n = -301/2 - √ 109561/4

Note that  √ 109561/4 can be written as
  √ 109561  / √ 4   which is 331 / 2

Solve Quadratic Equation using the Quadratic Formula

 3.3     Solving    n²+301n-4740 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  n  , the solution for   An²+Bn+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  n =   ————————
                      2A

  In our case,  A   =     1
                      B   =   301
                      C   =  -4740

Accordingly,  B²  -  4AC   =
                     90601 - (-18960) =
                     109561

Applying the quadratic formula :

               -301 ± √ 109561
   n  =    —————————
                          2

Can  √ 109561 be simplified ?

Yes!   The prime factorization of  109561   is
   331•331 
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).

√ 109561   =  √ 331•331   =
                ±  331 • √ 1   =
                ±  331

So now we are looking at:
           n  =  ( -301 ± 331) / 2

Two real solutions:

n =(-301+√109561)/2=(-301+331)/2= 15.000

or:

n =(-301-√109561)/2=(-301-331)/2= -316.000

Two solutions were found :

1.  n = 15
2.  n = -316