Q. if the square of difference of the zeroe of the quadratic polynomial f(x)=x^2=px=45 is equal to 144, then find the value of p 

ANS.  x^2+px+45=0                                                                         let roots be a and b  hence, sum of root (a+b)=p and product of roots(ab)=45

now,(a-b)^2=144

(a+b)^2-4ab=144

p^2-4*45=144

p^2=324

p=+18 and -18

in this answer how does 4ab comes

please answer fast