Write a Quine in Plain English

><>, 25 words + (22 + 1¹ + 1¹) extra = 57 49

Oh god, my brain.

'brr3deep*clap6beep+orb5flap*leap4deep+clap5beep5flap*leap9deep9clap*beep+flap0placed apple alp0leap=clip?lob8blip*flip0clip.

Since the rules state "There is no input.", for this program to work you'll need to pipe in an empty file.

Oh and yes, brr is a valid Scrabble word, as are tsktsk and pfft.

Explanation

First of all, the following words are no-ops:

deep clap beep flap leap clip blip flip

This is because of two reasons:

• abcdefl push numbers (10, 11, 12, 13, 14, 15 and the length of the stack respectively). In addition, i pushes -1 if EOF is met, seeing as there is no input.
• The p command pops three chars v,y,x and places v at the position x,y.

Yes, ><> lets you modify the source code on the fly! We don't really make use of that though, as we only need the p for popping.

If we get rid of these no-ops, we get:

'brr3*6+orb5*4+55*99*+0placed apple alp0=?lob8*0.

In a similar way, the laced app part of placed apple and the e alp part of apple alp are also no-ops, and the rr of brr just reverses the stack twice, which we can also remove:

'b3*6+orb5*4+55*99*+0pl0=?lob8*0.

Finally, something that looks like a regular ><> program! The idea is to make use of the standard ><> quine, which works like so:

• The initial ' starts string parsing, pushing every char met until we hit another '
• We keep pushing chars until the end of the line, at which point we wrap the instruction pointer back to the start (since ><> is toroidal)
• We land on the initial ' again, and stop string parsing. The result is that we've pushed every single char in the program (except the initial ') onto the stack.

Then the following happens:

b3*6+o                Print the initial ' quote (ASCII 39)
r                     Reverse the stack
b5*4+                 Push a ';' (ASCII 59)
55*99*+0p             Replace the 'l' of 'lob' with ';'

(print loop)
l0=?;                 If the stack is empty, terminate. Otherwise...
o                     Print the top of the stack
b8*0.                 Jump back to the beginning of the loop

golfscript, 8 words + 8 symbols = 20 16 (3?)

{four"words.written~twice"four}words.written~twice

The words are just a filler, the core is a tiny quine core:

{".~"}.~

Duplicates and evaluates a function that just appends the instructions to duplicate and evaluate itself. When a function is printed, it is stringified automatically. This is the smallest quine that does something.

Or we could just use a function literal that never gets evaluated. But it feels like cheating...

{whatever}

Chicken, 1 word

I'm not normally into esolangs, but this one seemed perfect for this. I think this is a true quine:

chicken

Pushes 1 chicken to the stack. The stack is then displayed.