Solution - Adding, subtracting and finding the least common multiple

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "2.7" was replaced by "(27/10)". 2 more similar replacement(s)

Step by step solution :

Step  1  :

            27
 Simplify   ——
            10

Equation at the end of step  1  :

    49        27
  ((——•(t2))-(——•t))-70  = 0 
    10        10
 Step  2  :
            49
 Simplify   ——
            10

Equation at the end of step  2  :

    49          27t     
  ((—— • t2) -  ———) -  70  = 0 
    10          10      

Step  3  :

Equation at the end of step  3  :

   49t2    27t     
  (———— -  ———) -  70  = 0 
    10     10      

Step  4  :

Adding fractions which have a common denominator :

 4.1       Adding fractions which 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)     49t2 - 27t
 ————————————  =  ——————————
      10              10    

Equation at the end of step  4  :

  (49t2 - 27t)    
  ———————————— -  70  = 0 
       10         

Step  5  :

Rewriting the whole as an Equivalent Fraction :

 5.1   Subtracting a whole from a fraction

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

          70     70 • 10
    70 =  ——  =  ———————
          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

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   49t² - 27t  =   t • (49t - 27) 

Adding fractions that have a common denominator :

 6.2       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:

 t • (49t-27) - (70 • 10)     49t2 - 27t - 700
 ————————————————————————  =  ————————————————
            10                       10       

Trying to factor by splitting the middle term

 6.3     Factoring  49t² - 27t - 700 

The first term is,  49t²  its coefficient is  49 .
The middle term is,  -27t  its coefficient is  -27 .
The last term, "the constant", is  -700 

Step-1 : Multiply the coefficient of the first term by the constant   49 • -700 = -34300 

Step-2 : Find two factors of  -34300  whose sum equals the coefficient of the middle term, which is   -27 .

For tidiness, printing of 30 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 - 27t - 700
  ————————————————  = 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:

  49t2-27t-700
  ———————————— • 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 :
   49t²-27t-700  = 0

Parabola, Finding the Vertex :

 7.2      Find the Vertex of   y = 49t²-27t-700

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, 49 , 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,At²+Bt+C,the  t -coordinate of the vertex is given by  -B/(2A) . In our case the  t  coordinate is   0.2755  

 Plugging into the parabola formula   0.2755  for  t  we can calculate the  y -coordinate : 
  y = 49.0 * 0.28 * 0.28 - 27.0 * 0.28 - 700.0
or   y = -703.719

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 49t²-27t-700
Axis of Symmetry (dashed)  {t}={ 0.28} 
Vertex at  {t,y} = { 0.28,-703.72} 
 t -Intercepts (Roots) :
Root 1 at  {t,y} = {-3.51, 0.00} 
Root 2 at  {t,y} = { 4.07, 0.00} 

Solve Quadratic Equation by Completing The Square

 7.3     Solving   49t²-27t-700 = 0 by Completing The Square .

 Divide both sides of the equation by  49  to have 1 as the coefficient of the first term :
   t²-(27/49)t-(100/7) = 0

Add  100/7  to both side of the equation :
   t²-(27/49)t = 100/7

Now the clever bit: Take the coefficient of  t , which is  27/49 , divide by two, giving  27/98 , and finally square it giving  729/9604 

Add  729/9604  to both sides of the equation :
  On the right hand side we have :
   100/7  +  729/9604   The common denominator of the two fractions is  9604   Adding  (137200/9604)+(729/9604)  gives  137929/9604 
  So adding to both sides we finally get :
   t²-(27/49)t+(729/9604) = 137929/9604

Adding  729/9604  has completed the left hand side into a perfect square :
   t²-(27/49)t+(729/9604)  =
   (t-(27/98)) • (t-(27/98))  =
  (t-(27/98))²
Things which are equal to the same thing are also equal to one another. Since
   t²-(27/49)t+(729/9604) = 137929/9604 and
   t²-(27/49)t+(729/9604) = (t-(27/98))²
then, according to the law of transitivity,
   (t-(27/98))² = 137929/9604

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-(27/98))²   is
   (t-(27/98))^2/2 =
  (t-(27/98))¹ =
   t-(27/98)

Now, applying the Square Root Principle to  Eq. #7.3.1  we get:
   t-(27/98) = √ 137929/9604

Add  27/98  to both sides to obtain:
   t = 27/98 + √ 137929/9604

Since a square root has two values, one positive and the other negative
   t² - (27/49)t - (100/7) = 0
   has two solutions:
  t = 27/98 + √ 137929/9604
   or
  t = 27/98 - √ 137929/9604

Note that  √ 137929/9604 can be written as
  √ 137929  / √ 9604   which is √ 137929  / 98

Solve Quadratic Equation using the Quadratic Formula

 7.4     Solving    49t²-27t-700 = 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   =   -27
                      C   =  -700

Accordingly,  B²  -  4AC   =
                     729 - (-137200) =
                     137929

Applying the quadratic formula :

               27 ± √ 137929
   t  =    ————————
                        98

  √ 137929   , rounded to 4 decimal digits, is  371.3879
 So now we are looking at:
           t  =  ( 27 ±  371.388 ) / 98

Two real solutions:

 t =(27+√137929)/98= 4.065

or:

 t =(27-√137929)/98=-3.514

Two solutions were found :

1.  t =(27-√137929)/98=-3.514
2.  t =(27+√137929)/98= 4.065