Question 1096362
There are infinite answers, but I will give you one.
Quadratic functions always have "two zeros" that can be
different real numbers, or
the same real number, or
conjugate complex numbers.
Given all the zeros of a quadratic function,
there is only one quadratic function with {{{1}}} as its "leading coefficient"
and exactly those zeros.
However, multiplying that function times any non-zero number,
You get another quadratic function with exactly the same zeros.
If {{{red((1-sqrt(3)/2))}}} and {{{green((1+sqrt(3)/2))}}} are zeros of a quadratic function,
the function can be written as
{{{a*(x-red((1-sqrt(3)/2)))*(x-green((1+sqrt(3)/2)))}}}={{{a*(x-1+sqrt(3)/2)*(x-1-sqrt(3)/2)}}}={{{a*((x-1)^2-3/4)}}}={{{a*(x^2-2x+1-3/4)}}}={{{a*(x^2-2x+1/4)}}}
So, one solution is {{{highlight(f(x)=x^2-2x+1/4)}}} .