In x^2+7x+4/49, the discriminant is 7^2-4*1*(4/49)=(49-16/49)/49=(49^2-16^2)/49=((49-16)(49+16))/49=(33xx65)/49=2145/49, though positive, is not the square of a rational number. Hence we cannot factorize it by splitting middle term.
Hence, the way is to find out zeros of quadratic trinomial x^2+7x+4/49. Zeros of ax^2+bx+c are given by quadratic formula (-b+-sqrt(b^2-4ac))/(2a).
So its zeros, which are two conjugate irrational numbers are given by quadratic formula and are
(-7+-sqrt(2145/49))/2 or
(-7+-sqrt2145/7)/2 or
-7/2+-sqrt2145/14 i.e. -7/2-sqrt2145/14 and -7/2+sqrt2145/14
Now, if alpha and beta are zeros of quadratic polynomial, then its factors are (x-alpha)(x-beta)
Hence factors of x^2+7x+4/49 are (x+7/2+sqrt2145/14) and (x+7/2-sqrt2145/14) and
x^2+7x+4/49=(x+7/2+sqrt2145/14)(x+7/2-sqrt2145/14)
