How many and what type of solutions does the equation have?
2c^2=16c−32
a. one rational solution
b. two nonreal solutions
c. two irrational solutions
d. two rational solutions

The answer is:  [A]:  "one rational solution" .

Explanation:

Given:  2c² = 16c − 32 ;  

Let us rewrite this equation into:

  "quadratic format" ; or:  "ax² + bx + c = 0 ;  a ≠ 0 " 


Given:  "2c² = 16c − 32" ; rewrite — substituting "x" for "c" ;

             →  2x² = 16x − 32 ;

             → Subtract "16x" ; and add "32" ; to EACH side of the equation;

             →  2x² − 16x + 32 = 16x − 32 − 16x + 32 ;  and rewrite:

             →  2x² − 16x + 32 = 0

So, this equation is now in "quadratic format" ; that is:

             →  ax² + bx + c = 0 ;  (a ≠ 0) ;
    
           in which:  a = 2 ;
                           b = -16 ;
                           c = 32 ;

So, we have the quadratic equation:

             →  2x² − 16x + 32 = 0 ;  
 
             →  Let us divide the ENTIRE EQUATION by "2" ;


             →  {2x² − 16x + 32} / 2 = 0 / 2 ;               
             
             →     x² − 8x + 16 = 0 ;
 
             → We can use the quadratic equation formula; but if we can factor the equation; it would be easier.  Can we factor it?  Yes!

        
            →     x² − 8x + 16  = (x − 4) (x − 4) ;

            →    (x − 4)(x − 4) = 0 ;

               Since each "multiplicand" is the same; that is: "(x − 4)" ; AND anything multiplied by "0" is equal to "0" ;  we can solve for "x" as follows:

            →  (x − 4) = 0 ;   Solve for "x" ;

            →   x − 4  = 0 ;   Add "4" to each side of the equation; to isolate "x" on one side of the equation; and to solve for "x" ;

            →   x − 4 + 4 = 0 + 4 ; 

            →   x = 4 .              

{Note: this is the long version; with practice (or inherently), by visual inspection, see that given: (x − 4) = 0 ; x = 4 .}.

So, since we substituted "x" for "c";  we rewrite the answer as:

           →  c = 4 .

 Let us check our

            By plugging in "4" for "c" ; in the original equation:

            →  2c² = 16c − 32  ;

            →  2*4² =? 16*4  − 32  ?? :
  
            →  2*16 =?  64 − 32 ??
              
            →   32   =   32 .  Yes!

         Since there is only one value for "c" ; which is, "4" (positive 4), the answer to the question is:  Answer choice: [A]: "one rational solution" .