Hint $\rm\:f(x) = ax^2\!+\!bx\!+\!c,\ a\ne 0\:$ and $\rm\:f(x) = f(y) = f(z)\:$ implies the following determinant $= 0$, having proportional first and last columns
$\rm 0\, =\, \left | \begin{array}{ccc} \rm f(x) & \rm x & 1 \\ \rm f(y) & \rm y & 1 \\ \rm f(z) & \rm z & 1 \end{array} \right | \, =\, \left | \begin{array}{ccc} \rm ax^2 & \rm x & 1 \\ \rm ay^2 & \rm y & 1 \\ \rm az^2 & \rm z & 1 \end{array} \right | \, =\, a\,(x-y)(y - z)(z-x) $
Thus, if the coefficient ring is a field (or domain), one of the RHS factors must be $0$. Since $\rm\:a\ne 0\:$ one of the other factors $= 0,\:$ i.e. the roots are not distinct.
Remark $ $ That the second determinant equals the first follows from the fact that the first columns are congruent modulo the others. More precisely, the second det arises by applying the elementary col operation $\rm\: col_1 \leftarrow col_1 - b\cdot col_2 - c\cdot col_3,\:$ yielding $\rm\: f(x) - b\cdot x - c\cdot 1 = ax^2,\:$ etc. The same argument works for higher degree polynomials. For the general case, it helps to know that the rhs matrix, after removing the factor of $\rm\:a,\:$ is the ubiquitous Vandermonde matrix,