#### Answer

$x_{1}=5$
$x_{2}=-13$

#### Work Step by Step

As $g(x)$ is given, this is the output value therefore we have to calculate the input value of $x$.
$g(x)=4x^2+32x-13$
$247=4x^2+32x-13$
$0=4x^2+32x-260$
$0=x^2+8x-65$
We can now substitute the equation into the quadratic formula to solve it.
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-8\pm\sqrt{8^2-4(1)(-65)}}{2(1)}=\frac{-8\pm18}{2}=-4\pm9$
The two solutions of the equation are:
$x_{1}=5$
$x_{2}=-13$