Cool Quadratic

Let, $\sqrt{x^2+1}=p .$

Square both side, $x^2+1=p^2$ (with $p>0).$ substitute both equation to original problem, $p+p^2=90 \implies p^2+p-90=0 \implies (p+10)(p-9)=0 \\ \implies p=9$ (because $p=-10 <0).$

subtitute $p=9$ to the second equation ${x^2+1}=9^2 \implies x= \pm \sqrt{80} .$ sum of the roots is $(- \sqrt{80})+( \sqrt{80})= \boxed 0$

if (a) satisfy the equation we note that (-a) satisfy the equation then the sum of all values of which satisfy the equation = 0

assume, x² + 1 = t² ............... =>t = 9 ............. => x = ±sqrt(80)

I think it is plain that if we have only $x^{2}$ terms in an equation it suggests if x is a root,so is -x .We needn't check for real solutions because the roots can be complex too(it hasn't been suggested anywhere that we ignore the complex roots).Hence the answer without any computation yields zero

If we take x^2 +1 to the other side and square both sides we will get a biquadratic equation .here the coefficent of x^3 is 0 so the sum of roots is 0

${ x }^{ 2 }+1=n$

Then,

$\sqrt { n } +n=90\\ \sqrt { n } =90-n\\ n={ n }^{ 2 }-180n+8100\\ { n }^{ 2 }-181n+8100=0\\ { \left( { x }^{ 2 }+1 \right) }^{ 2 }-181\left( { x }^{ 2 }+1 \right) +8100=0\\ { x }^{ 4 }+2{ x }^{ 2 }+1-181{ x }^{ 2 }-181+8100=0\\ { x }^{ 4 }-179{x}^{2}+7920=0$

Using vieta's , we get that the sum of the roots is $-\frac { 0 }{ 1 } =0$

Alternatively, you could also have exploited the fact that there will be a ${ x }^{ 4 }$ term, but there will not be a ${ x }^{ 3 }$ term in the quartic.