If the roots of $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c$ is

Consider the following two statements:
$I$. Any pair of consistent linear equations in two variables must have a unique solution.
$II$. There do not exist two consecutive integers,the sum of whose squares is $365$.
Then,

If $a, b, c, d$ are real numbers such that $a < b < c < d$,then the roots of the equation $(x-a)(x-c)+2(x-b)(x-d)=0$ are

If $x = 2 + 2^{2/3} + 2^{1/3},$ then $x^3 - 6x^2 + 6x = $

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

If $f(x) = ax^{2} + bx + 2$ and $f(1) = 4, f(3) = 38$,then $a - b = $