Question 1006597
There's no need to use the quadratic formula since that would require us to expand/simplify the left side. It's more work than needed in my opinion.



Since it's in vertex form, we can isolate x to get



{{{4(x-5)^2 -1=0}}}



{{{4(x-5)^2 -1+1=0+1}}} Add 1 to both sides



{{{4(x-5)^2=1}}}



{{{(1/4)*4(x-5)^2=(1/4)*1}}} Multiply both sides by 1/4



{{{(x-5)^2=1/4}}}



{{{sqrt((x-5)^2)=sqrt(1/4)}}} Apply the square root to both sides



{{{abs(x-5)=sqrt(1/4)}}} Use the rule {{{sqrt(x^2) = abs(x)}}} where x is any real number



{{{x-5=1/2}}} or {{{x-5=-1/2}}} Split up the absolute value into the plus/minus components



{{{x-5+5=1/2+5}}} or {{{x-5+5=-1/2+5}}} For each equation, add 5 to both sides



{{{x = 1/2+10/2}}} or {{{x = -1/2+10/2}}}



{{{x = 11/2}}} or {{{x = 9/2}}}



-------------------------------------------



So the two solutions to {{{4(x-5)^2 -1=0}}} are...



{{{x = 11/2}}} or {{{x = 9/2}}}


------------------------------------------



So g(x) is undefined when {{{x = 11/2}}} or {{{x = 9/2}}}


Side Notes: 
{{{4 & 1/2 = 9/2}}}
{{{5 & 1/2 = 11/2}}}