format long % show all the digits
a = 1; b = -(1e6+1e-6); c = 1;
x1 = (-b + sqrt(b^2-4*a*c)) / (2*a)
x2 = (-b - sqrt(b^2-4*a*c)) / (2*a)
The first value is correct to all stored digits, but the second has fewer than six accurate digits:
-log10( abs(1e-6-x2)/1e-6 )