Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "2.4" was replaced by "(24/10)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

            12
 Simplify   ——
            5 

Equation at the end of step  1  :

       49             12
  ((0-(——•(t2)))+27t)+——  = 0 
       10             5 
 Step  2  :
            49
 Simplify   ——
            10

Equation at the end of step  2  :

          49                   12
  ((0 -  (—— • t2)) +  27t) +  ——  = 0 
          10                   5 

Step  3  :

Equation at the end of step  3  :

         49t2             12
  ((0 -  ————) +  27t) +  ——  = 0 
          10              5 

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  10  as the denominator :

           27t     27t • 10
    27t =  ———  =  ————————
            1         10   

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 4.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 -49t2 + 27t • 10     270t - 49t2
 ————————————————  =  ———————————
        10                10     

Equation at the end of step  4  :

  (270t - 49t2)    12
  ————————————— +  ——  = 0 
       10          5 

Step  5  :

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   270t - 49t²  =   -t • (49t - 270) 

Calculating the Least Common Multiple :

 6.2    Find the Least Common Multiple

      The left denominator is :       10 

      The right denominator is :       5 

      Least Common Multiple:
      10 

Calculating Multipliers :

 6.3    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = 2

Making Equivalent Fractions :

 6.4      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)²   and  (y²+y)/(y+1)³  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.      -t • (49t-270)
   ——————————————————  =   ——————————————
         L.C.M                   10      

   R. Mult. • R. Num.      12 • 2
   ——————————————————  =   ——————
         L.C.M               10  

Adding fractions that have a common denominator :

 6.5       Adding up the two equivalent fractions

 -t • (49t-270) + 12 • 2     -49t2 + 270t + 24
 ———————————————————————  =  —————————————————
           10                       10        

Trying to factor by splitting the middle term

 6.6     Factoring  -49t² + 270t + 24 

The first term is,  -49t²  its coefficient is  -49 .
The middle term is,  +270t  its coefficient is  270 .
The last term, "the constant", is  +24 

Step-1 : Multiply the coefficient of the first term by the constant   -49 • 24 = -1176 

Step-2 : Find two factors of  -1176  whose sum equals the coefficient of the middle term, which is   270 .

For tidiness, printing of 18 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step  6  :

  -49t2 + 270t + 24
  —————————————————  = 0 
         10        

Step  7  :

When a fraction equals zero :

 7.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  -49t2+270t+24
  ————————————— • 10 = 0 • 10
       10      

Now, on the left hand side, the  10  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -49t²+270t+24  = 0

Parabola, Finding the Vertex :

 7.2      Find the Vertex of   y = -49t²+270t+24

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -49 , is negative (smaller 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,At²+Bt+C,the  t -coordinate of the vertex is given by  -B/(2A) . In our case the  t  coordinate is   2.7551  

 Plugging into the parabola formula   2.7551  for  t  we can calculate the  y -coordinate : 
  y = -49.0 * 2.76 * 2.76 + 270.0 * 2.76 + 24.0
or   y = 395.939

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -49t²+270t+24
Axis of Symmetry (dashed)  {t}={ 2.76} 
Vertex at  {t,y} = { 2.76,395.94} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = { 5.60, 0.00} 
Root 2 at  {t,y} = {-0.09, 0.00} 

Solve Quadratic Equation by Completing The Square

 7.3     Solving   -49t²+270t+24 = 0 by Completing The Square .

 Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
 49t²-270t-24 = 0  Divide both sides of the equation by  49  to have 1 as the coefficient of the first term :
   t²-(270/49)t-(24/49) = 0

Add  24/49  to both side of the equation :
   t²-(270/49)t = 24/49

Now the clever bit: Take the coefficient of  t , which is  270/49 , divide by two, giving  135/49 , and finally square it giving  18225/2401 

Add  18225/2401  to both sides of the equation :
  On the right hand side we have :
   24/49  +  18225/2401   The common denominator of the two fractions is  2401   Adding  (1176/2401)+(18225/2401)  gives  19401/2401 
  So adding to both sides we finally get :
   t²-(270/49)t+(18225/2401) = 19401/2401

Adding  18225/2401  has completed the left hand side into a perfect square :
   t²-(270/49)t+(18225/2401)  =
   (t-(135/49)) • (t-(135/49))  =
  (t-(135/49))²
Things which are equal to the same thing are also equal to one another. Since
   t²-(270/49)t+(18225/2401) = 19401/2401 and
   t²-(270/49)t+(18225/2401) = (t-(135/49))²
then, according to the law of transitivity,
   (t-(135/49))² = 19401/2401

We'll refer to this Equation as  Eq. #7.3.1  

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

Note that the square root of
   (t-(135/49))²   is
   (t-(135/49))^2/2 =
  (t-(135/49))¹ =
   t-(135/49)

Now, applying the Square Root Principle to  Eq. #7.3.1  we get:
   t-(135/49) = √ 19401/2401

Add  135/49  to both sides to obtain:
   t = 135/49 + √ 19401/2401

Since a square root has two values, one positive and the other negative
   t² - (270/49)t - (24/49) = 0
   has two solutions:
  t = 135/49 + √ 19401/2401
   or
  t = 135/49 - √ 19401/2401

Note that  √ 19401/2401 can be written as
  √ 19401  / √ 2401   which is √ 19401  / 49

Solve Quadratic Equation using the Quadratic Formula

 7.4     Solving    -49t²+270t+24 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  t  , the solution for   At²+Bt+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  t =   ————————
                      2A

  In our case,  A   =    -49
                      B   =   270
                      C   =   24

Accordingly,  B²  -  4AC   =
                     72900 - (-4704) =
                     77604

Applying the quadratic formula :

               -270 ± √ 77604
   t  =    ————————
                        -98

Can  √ 77604 be simplified ?

Yes!   The prime factorization of  77604   is
   2•2•3•29•223 
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).

√ 77604   =  √ 2•2•3•29•223   =
                ±  2 • √ 19401

  √ 19401   , rounded to 4 decimal digits, is  139.2875
 So now we are looking at:
           t  =  ( -270 ± 2 •  139.287 ) / -98

Two real solutions:

 t =(-270+√77604)/-98=(135-√ 19401 )/49= -0.087

or:

 t =(-270-√77604)/-98=(135+√ 19401 )/49= 5.598

Two solutions were found :

1.  t =(-270-√77604)/-98=(135+√ 19401 )/49= 5.598
2.  t =(-270+√77604)/-98=(135-√ 19401 )/49= -0.087