Two of its zeroes of the quartic equation are known values 2+\sqrt{3} and 2-\sqrt{3}. Let a,b be another two zeroes of the equation.

According to Vieta's formula, we have the following two equations

a+b+2+\sqrt{3}+2-\sqrt{3}= 6

ab(2+\sqrt{3})(2-\sqrt{3}) = -35

Simplifying gives the system of equations

\begin{cases} a + b=2 \\ ab=-35 \end{cases}

Applying Vieta's formula shows a,b are two roots of the quadratic equation

x^2-2x-35=0

Solve for x

x=7 or x=-3

Therefore, the other two roots for degree 4 polynomial are 7 and -3