Since the longest side must be opposite the largest angle in the triangle, the side which is 28 units is opposite the angle of degree measure of 120. Let us name the remaining sides $a$ and $b$. Using the cosine rule we get: $28^2=a^2+b^2-2ab\cos(120)$ $784=a^2+b^2-2ab\left(\frac{-1}{2}\right)$ $784=a^2+b^2+ab$ Since $ab=240$ $a^2+b^2=544$ We thus have the following system of equations to solve: $\begin{align*}a^2+b^2&=544 \\ ab&=240\end{align*}$ Solving for $b$ in the latter equation $b=\frac{240}{a}$ Plugging into the first one gives: $a^2+\left(\frac{240}{a}\right)^2=544$ Multiplying both sides by $a^2$ and letting $u=a^2$ $u^2-544u+240^2=0$ Using the quadratic equation $\begin{align*}u&=\frac{544\pm\sqrt{(-544)^2-4\cdot240^2}}{2} \\ u&=\frac{544\pm256}{2} \\ u&=144 \text{ or }400\end{align*}$ But since $u=a^2$, $a^2=144 \text{ or } a^2=400$ $a=12 \text{ or } a=20$ Using $b=\dfrac{240}{a}$ we thus get $a=12$ and $b=20$ or $a=20$ and $b=12$. In either case the sum of the sides is 60.
Edit: lhf pointed out a much quicker alternative:
When we came to the following stage $784=a^2+b^2+ab$ Rather than solve for $a$ and $b$ etc. simply add $ab$ to both sides to get $784+ab=a^2+b^2+2ab$ Since $a^2+b^2+2ab$ is a perfect square and $ab=240$ $1024=(a+b)^2$ $a+b=32$ Therefore the sum of all the sides is $a+b+28=32+28=60$