You have to try to solve the associated equation:
#ax^2+bx+c=0#
and if the #Delta=b^2-4ac# is positive it could be factored:
#a(x-x_1)(x-x_2)# where #x_1 and x_2# are the two solution:
#x_(1,2)=(-b+-sqrtDelta)/(2a)#.
If #Delta# is zero, than it it a square of a binomial, and if #Delta# is negative, it couldn't be factored.
So, in our case:
#Delta=7^2-4*2*5=49-40=9#, so we can find the two solutions:
#x_(1,2)=(-7+-3)/4#
and so:
#x_1=(-7-3)/4=-10/2=-5/2#
#x_2=(-7+3)/4=-4/4=-1#
it could be factored:
#2(x+5/2)(x+1)# or #(2x+5)(x+1)#.
