Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

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