If equation is $ax^2 + bx+c=0$ and roots of the equation are $x_1$ and $x_2$
then $x_1 + x_2 = -b/a$
and $x_1 . x_2 =c/a$
in 1st attempt $c$ is incorrect, So, we can say $x_1 + x_2 = 4+3=7\quad \to(i)$ [as here $-b/a$ is correct]
in 2nd attempt $b$ is incorrect. So, we can say $x_1 . x_2 =3.2=6\quad \to (ii)$ [as here $c/a$ is correct]
Now solving $(ii)$ we get,
$x_2 =6 / x_1\quad \to (iii)$
Putting it in eqn. $(i)$
$x_1 + \frac{6}{ x_1} = 7$
$\implies x_1^2 -7x_1 +6 =0$
$\implies x_1 =1$ or $x_1 = 6$
Answer is (C)