As noticed by Sasha in his comment, the answer is no.
Moreover, the fact that the degree is $2$ is very important here. In fact, 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 $X_d$ is equal to
$\chi(X_d)= \langle c_n(T_{X_d}), [X_d] \rangle =\frac{1}{d}[(1-d)^{n+2}-1]+n+2$,
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 results is Dimca's book "Singularities and topology of hypersurfaces", Chapter 5, which also considers the case of hypersurfaces in $\mathbb{P}^{n+1}$ with isolated singularities.