$\phi(n)=(p-1)\times(q-1)$

$n=pq$

$2880=pq-p-q+1$

$128=p+q...................(1)$

$3007=pq.....................(2)$

$Using\ (1)\ \&\ (2)$

$q^2-128q+3007$

$Roots:$

$\\ \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ \dfrac{128\pm \sqrt{(128)^2-4\times1\times3007}}{2}\\ \\ \dfrac{128\pm66}{2}\\ \\ \dfrac{194}{2}=97\checkmark\ \&\ \dfrac{62}{2}=31$