\(ax^2+bx+c=0\ have\ two\ roots\ if\ b^2-4ac > 0.\\-------------------------\\x^2+kx+3k-5=0\\\\a=1;\ b=k;\ c=3k-5\\\\b^2-4ac=k^2-4\cdot1\cdot(3k-5)=k^2-12k+20 > 0\\\\k^2-2k-10k+20 > 0\\\\k(k-2)-10(k-2) > 0\\\\(k-2)(k-10) > 0\\\\k=2;\ k=10\ (look\ at\ the\ picture)\\\\k\in(-\infty;\ 2)\ \cup\ (10;\ \infty)\)
