+580 I'm sorry to say this but, think again guest answerer 
\(\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)
(*ahhhh man not this again...)
Just know that:
\(a=2y,\:\:b=-7\)
when we are substituting values in the formula!!!
\(=\sum _{i=0}^5\binom{5}{i}\left(2y\right)^{\left(5-i\right)}\left(-7\right)^i\)
\(=\frac{5!}{0!\left(5-0\right)!}\left(2y\right)^5\left(-7\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2y\right)^4\left(-7\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2y\right)^3\left(-7\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2y\right)^2\left(-7\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2y\right)^1\left(-7\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2y\right)^0\left(-7\right)^5\)
\(32y^5-560y^4+3920y^3-13720y^2+24010y-16807\)
The coefficient of y^4 in the expansion of (2y-7)^5 is \(-560\)
-Vinculum
!
