We can use the identity #a^2-b^2=(a+b)(a-b)#.
Note that #v^2+14v# can be converted to a square by adding square of half of the coeffucient of #x#. As coefficient of #x# ia #14#, square of half of #14# is #7^2=49# and then we can write it as #(v+7)^2# as #(x+a)^2=x^2+2ax+a^2#. Hence
#v^2+14v-3=0# can be written as
#v^2+14v+49-49-3=0#
or #(v+7)^2-52=0#
or #(v+7)^2-(sqrt52)^2=0#
or #(v+7)^2-(2sqrt13)^2=0#
and now using #a^2-b^2=(a+b)(a-b)#, we can write it as
#(v+7+2sqrt13)(v+7-2sqrt13)=0#
Although factorization is over, moving further, we can say
#(v+7+2sqrt13)(v+7-2sqrt13)=0# means
either #v+7+2sqrt13=0# or #v+7-2sqrt13=0#
i.e. #v=-7-2sqrt13# or #v=-7+2sqrt13#
