Find the solution set of 7q^2-28=0

The solution set is:  q = {2, -2) ;   [ or, write as:  q = {± 2}  ] .

Explanation:

GIven:

   7q²  −  28  =  0  ;

   Divide the entire equation by "7" (both sides), to simplify;

          →   (7q²  −  28) / 7 = 0 / 7 ; 

 To get:     →   q²  − 4 = 0 ;

We can factor:  " q²  − 4  " ;  into:  "(q +2)*(q− 2)" ; 

 and rewrite:  q²  − 4 = 0 ; 

 as :     (q +2)(q− 2) = 0 ;

  So, since anything multiplied by "0" equals "0" ; 
 
 When "(q+2)" = 0 ; the equation  holds true. 

So,  q + 2 = 0 ;

→ Subtract "2" from EACH side of the equation:

    q + 2 − 2  = 0 − 2 ;

 to get:  q = -2 ; 

 As such, when "(q−2)" = 0, the equation holds true. 

 So,   q − 2  = 0 ;

Add "2" to both sides of the equation:

         q − 2 + 2  = 0 + 2 ;

 to get:   q = 2 .

So we have 2 (TWO) values for "q" —which are "2" and "-2" .

      So the solution set is:  q = {2, -2) ;   [or, write as:  q = {± 2}  ] .

The answer to your question is 1/9