Finding x using the pythagoras theorem $$x^2 = (x+1)^2 + (x-7)^2$$
can someone please find $x$? Also this is a quadratic equation problem solving question. 

As in user$169852$'s answer
$$x^2 = (x+1)^2 + (x-7)^2 \quad
\implies x^2 - 12 x + 50 = 0$$
and the quadratic formula shows that $x$ is complex.
$$x=\dfrac{12\pm\sqrt{144-4(1)50}}{2(1)}
=
\dfrac{12\pm2i\sqrt{14}}{2} 
= 6\pm i\sqrt{14}
$$
however by the same logic
$$ x^2 = (x+1)^2 + (x+7)^2\quad  \implies 
x=-8\pm \sqrt{14}$$
and
$$x^2 = (x+1)^2 - (x - 7)^\quad \implies x\in\{4,12\}$$
I suspect that the latter is what you are looking for where
$A= x-7=5\qquad  B=x=12\qquad C=x+1=13
$
$$\implies (A,B,C)=(5,12,13)$$
