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Francesco Polizzi
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As noticed by Sasha in his comment, the answer is no and the fact that the degree is $2$ is very important here.

In general, given a smooth complex hypersurface of degree d, say $X_d \subset \mathbb{P}^{n+1}$, by standard arguments involving Lefschetz theorem we have

$H^k(X_d)= H^k(\mathbb{P}^n)$ for $k \neq n$.

In particular, all the odd Betti numbers are zero, except possibly the middle Betti number when $n$ is odd. On the other hand, the Euler-Poincare characteristic of $X_d$ can be computed and it is equal to

$\chi(X_d)=\frac{1}{d}[(1-d)^{n+2}-1]+n+2$,

so for $n$ odd and $d=2$ the middle cohomology group must be zero too. Notice that for $n$ odd and $d >2$ one always has a non-zero middle Betti number. For instence, if $X \subset \mathbb{P}^4$ is a smooth cubic hypersurface, then $b_3(X)=4$.

A good reference on these topics is Dimca's book "Singularity and topology of hypersurfaces".

Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283