Consider the quadratic equation in standard form

ax² + bx + c = 0

The roots are

x = (−b ±(b² − 4ac)^½)/2a
or, equivalently
x = (−b ± b(1 − 4ac/b²)^½)/2a

Note that for small z

(1 + z)½ ≅ 1 + z/2

Therefore

x ≅ (−b ± b(1 − 2ac/b²) /2a

Consider first the positive value of the radical; i.e.'

x ≅ (−b + b − 2ac/b) /2a
which reduces to
x ≅ − (2ac/b)/2a = − c/b

Thus for small values of c, x=−c/b regardless of the magnitude of a.

For the negative value of the radical

x = (−b −b(1 − 4ac/b²)/2a
and hence
x ≅ (−2b + 2ac/b)/2a = −(b/a) + (c/b)