Suppose I have a smooth 2dimensional quadric bundle $f:X\to S$ over a surface $S$. Suppose furthermore that the discriminant locus $\Delta \subset S$ is smooth. Can I immedately conclude that the fibers of $f$ have at most isolated singularities? Why?

It is sufficient to prove that if the fibre over a point $s \in S$ is the union of two planes or a plane counted twice, then $\Delta$ is singular at $s$. The question being local, we may assume that $S$ is a small polidisk centered at $(0,0) \in \mathbb{C}^2$ and $s=(0,0)$. Let $\mathcal{O}$ be the local ring of convergent power series centered in $(0,0)$ and $\mathfrak{m} \subset \mathcal{O}$ be its maximal ideal, namely the set of power series vanishing at $(0,0)$. In the case where the central fibre is the union of two planes, the equation of our conic bundle is $$2x_0x_1 + \sum a_{ij}(z,w)x_i x_j=0,$$ where $i,j \in \{0,1,2,3\}$, $\{i,j \} \neq \{ 0,1 \}$ and $a_{ij}(z,w)=a_{ji}(z,w) \in \mathfrak{m} $. In fact, the fibre over $(0,0)$ is the reducible quadric $x_0x_1=0$. The equation of the discriminant $\Delta$ is then given by $$\det \left(\begin{array}{llll} a_{00} & 1 & a_{02} & a_{03} \\\ 1 & a_{11} & a_{12} & a_{13} \\\ a_{02} & a_{12} & a_{22} & a_{23} \\\ a_{03} & a_{13} & a_{23} & a_{33} \end{array}\right)=0.$$ Since $a_{ij} \in \mathfrak{m}$, the explicit computation shows that the determinant above belongs to $\mathfrak{m}^2$. This precisely means that $\Delta$ is singular at $(0,0)$, proving the claim. In the case where the central fibre is a plane counted twice, the equation of our conic bundle is $$x_0^2 + \sum a_{ij}(z,w)x_i x_j=0,$$ with the $a_{ij}(z,w)$ as above. In fact, the fibre over $(0,0)$ is the double plane $x_0^2=0$. The proof that $\Delta$ is singular at $(0,0)$ is the same as in the previous case. 

