There can be no solution with 32 coins.
Perhaps you meant 33 coins instead.  I will
assume so.
Let the number of quarters be x
Let the number of dimes be y
                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
quarters     x        $0.25     $0.25x
dimes        y        $0.10     $0.10y
-------------------------------------------
TOTALS      33      -----       $5.25
The first equation comes from the "Number of coins" column.
  



                 x + y = 33
 The second equation comes from the last column.
  



         0.25x + 0.10y = 5.25
Get rid of decimals by multiplying every term by 100:
             25x + 10y = 525
 So we have the system of equations:
           
.
We solve by substitution.  Solve the first equation for y:
           x + y = 33
               y = 33 - x
Substitute (33 - x) for y in 25x + 10y = 525
   25x + 10(33 - x) = 525
    25x + 330 - 10x = 525
          15x + 330 = 525
                15x = 195
                  x = 13 = the number of quarters.
Substitute in y = 33 - x
              y = 33 - (13
              y = 20 dimes.
Checking:  13 quarters is $3.25 and 20 dimes is $2.00
            That's 33 coins.
            And indeed $3.25 + $2.00 = $5.25
Edwin