Question 957082
{{{f(x)=x+1}}}
{{{x+1=17}}}
{{{x=16}}}
So then,
{{{g(x)=16}}}
{{{2x^2=16}}}
{{{x^2=8}}}
{{{x= 0 +- sqrt(8)}}}
{{{x= 0 +- 2sqrt(2)}}}