Statement -I : The value should be subtracted from $x^2-16x+30$ to get 15 as one of its zeroes is 15.

Statement -II : The common root of the equations $x^2-bx+c=0$ and $x^2+bx-a=0$ is $\frac{c+a}{2b}$

Take $(x^2-16x+30)-15=x^2-16x+15$$=(x-15)(x-1)$

$\Rightarrow x=15$ or $x=1$

$\therefore \mathrm{statement} \left(\mathrm{A}\right) \mathrm{is} \mathrm{TRUE}$

Statement (B)

Take $x^2-bx+c=0$

$\frac{(c-a{)}^2}{4b^2}-\frac{b-(c+a)}{2b}+c=\frac{(c+a{)}^2-2b^2(c+a)+4b^{2}c}{4b^2}$

$=\frac{c^2+a^2+2ac-2b^{2}c-2b^{2}a+4b^{2}c}{4b^2}$

$=\frac{c^2+a^2+2ac-2b^{2}a+2b^{2}c}{4b^2}\to \left(i\right)$

Take $x^2+bx-a=0$

$\Rightarrow \frac{(c+a{)}^2}{4b^2}+\frac{b(c+a)}{2b}-a=\frac{(c+a{)}^2+2b^2(c+a)-4a{b}^2}{4b^2}\to \left(ii\right)$

$\left(i\right)=\left(ii\right)\Rightarrow$statement (B) is TRUE